Question 12 Marks
A die, whose faces are marked 1, 2, 3 in red and 4, 5, 6 in green, is tossed. Let A be the event ‘‘number obtained is even’’ and B be the event ‘‘number obtained is red’’. Find if A and B are independent events.
AnswerEvent A: Number obtained is even
B: Number obtained is red.
$\text{P(A)} = \frac{3}{6} = \frac{1}{2}, \text{P(B)} = \frac{3}{6} = \frac{1}{2}$
$\text{P(A} \cap \text{B}) = \text{P}$ (getting an even red number) $= \frac{1}{6}$
$\text{Since P(A).P(B)} = \frac{1}{2}.\frac{1}{2} = \frac{1}{4} \neq \text{P(P}\cap\text{B)} \text{ which is } \frac{1}{6}$
$\therefore$ A and B are not independent events.
View full question & answer→Question 22 Marks
$\text{If P(A) = 0·4, P(B) = p, P(A}\cup \text{B) = 0·6}$ and A and B are given to be independent events, find the value of ‘p’.
Answer$\text{ P(A} \cup \text{B}) = \text{P(A)} + \text{P(B) - P(A} \cap \text{B})$
$\text{= P(A) + P(B) – P(A) P(B)}$ as A and B are independent events.
$\therefore \text{ 0.6 = 0.4 + p – (0.4)p}$
$\Rightarrow \text{p} = \frac{1}{3}$
View full question & answer→Question 32 Marks
Prove that if E and F are independent events, then the events E and F' are also independent.
Answer$\text{P(E } \cap \text {F}') = \text{P(E) - P(E} \cap \text{F)}$
$= \text{P(E) - P (E) . P(F))}$
$= \text{P(E)[1 - P(F)]}$
$= \text{P(E) P(F}')$
$\Rightarrow \text{E and F}'$ are independent events.
View full question & answer→Question 42 Marks
A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
AnswerLet F : number of red die is less than 4.
E : sum of number is 8
E = {(2, 6)(3, 5)(4, 4)(5, 6)(6, 2)}
$\Rightarrow\text{P}(\text{E})=\frac{5}{36}$
$\text{F}= \{(1, 1)(2, 1)(3, 1) ........ (6, 1)$
$(1, 2)(2, 2) ............. (6, 2)$
$(1, 3)(2, 3) ............... (6, 3)\}$
$\text{P}(\text{E})=\frac{18}{3 6}$
Also $\text{E}\cap\text{F}=\{(5,3)(6,2)\}\Rightarrow\text{P}(\text{E}\cap\text{F})=\frac{2}{36}$
$\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)=\frac{\text{P}(\text{E}\cap\text{F})}{\text{P}(\text{F})}=\frac{\frac{2}{36}}{\frac{18}{36}}$
$=\frac{1}{9}$
View full question & answer→Question 52 Marks
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event “number is even” and B be the event “number is marked red”. Find whether the events A and B are independent or not.
AnswerWhen a die is thrown, the sample space (S) is
S = {1, 2, 3, 4, 5, 6}
Let A: the number is even = {2, 4, 6}
B: the number is red = {1, 2, 3}
$\Rightarrow\text{P(B)}=\frac{3}{6}=\frac{1}{2}$
$\therefore\text{A}\cap\text{B}=\big\{{2\big\}}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
$\Rightarrow\text{P(A)}\times\text{P(B)}\neq\text{P}(\text{A}\cap\text{B})$
Therefore, A and B are not independent.
View full question & answer→Question 62 Marks
A die is thrown 6 times. If “getting an odd number” is a “success”, what is the probability of:
-
5 successes?
-
Atmost 5 successes?
AnswerLet's consider success as 'p' and failure as 'q'.
Here we have 3 odd and 3 even numbers out of total 6 sample spaces
$\text{P(p)}=\frac{1}{2}$ and $\text{P(q)}=\frac{1}{2}$
- Probability of getting 5 successes =
$^6\text{C}_5\Big(\frac{1}{2}\Big)^5\Big(\frac{1}{2}\Big)^1=\ ^6\text{C}_5\Big(\frac{1}{2}\Big)^6=\frac{3}{32}.$
- Probability of getting atmost 5 successes =
$\text{P}(\text{p}\leq5)=1-\text{P}(\text{p}=6)$
$=1-\ ^6\text{C}_6\Big(\frac{1}{2}\Big)^6=1-\frac{1}{64}$
$=\frac{63}{64}.$ View full question & answer→Question 72 Marks
The random variable X has a probability distribution P(X) of the following form, where ‘k’ is some number.
$\text{P}(\text{X}=\text{x})=\begin{cases}\text{k}, & \text{if x}=0\\2\text{k}, & \text{if x}=1\\3\text{k}, & \text{if x}=2\\0, & \text{otherwise}\end{cases}$
Determine the value of ‘k’.
AnswerIt is known that the sum of probabilities of a probability distribution of random variables is one.
$\therefore\text{k}+2\text{k}+3\text{k}+0=1$
$\Rightarrow6\text{k}=1$
$\Rightarrow\text{k}=\frac{1}{6}$
View full question & answer→Question 82 Marks
The probability of finding a green signal on a busy crossing X is 30%. What is the probability of finding a green signal on X on two consecutive days out of three?
AnswerGiven the probability of finding a green signal on a busy crossing X is 30%. What is the probability of finding a green signal on X on two consecutive days out of three?
Let S be the event of a green signal.
The probability of finding a green signal on a busy crossing is $30\%= \frac{3}{10}$
Therefore required probability is $\Big(\frac{3}{10}\Big)^2\times\frac{7}{10}+\frac{7}{10}\times\Big(\frac{3}{10}\Big)^2$
$= \frac{9}{100}\times\frac{7}{10}+\frac{7}{10}\times\frac{9}{100}$
$= \frac{63}{1000} + \frac{63}{1000}$
$=\frac{63+63}{1000}$
$=\frac{126}{1000}$
View full question & answer→Question 92 Marks
A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its cooler is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be,
Blue and red in any order.
AnswerBag contains 3 blue, 5 red marble. One marble is drawn, its colour noted and replaced then again a marble drawn and its colour is noted.
P (Blue and red in any order)
$=\text{P}\big((\text{B}\cap\text{R})\cup(\text{R}\cap\text{B})\big)$
$=\text{P}(\text{B}\cap\text{R})+\text{P}(\text{R}\cap\text{B})$
$=\text{P}(\text{B})\times\text{P}(\text{R})+\text{P}(\text{R})\times\text{P}(\text{B})$
$=\frac{3}{8}\times\frac{5}{8}+\frac{5}{8}\times\frac{3}{8}$
$=\frac{30}{64}$
$=\frac{15}{32}$
Required probability $=\frac{15}{32}$
View full question & answer→Question 102 Marks
A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its cooler is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be,
Blue followed by red.
AnswerBag contains 3 blue, 5 red marble. One marble is drawn, its colour noted and replaced then again a marble drawn and its colour is noted.
P (Blue followed by red)
$=\text{P}(\text{B}\cap\text{R})$
$=\text{P}(\text{B})\times\text{P}(\text{R})$
$=\frac{3}{8}\times\frac{5}{8}$
$=\frac{15}{64}$
Required probability $=\frac{15}{64}$
View full question & answer→Question 112 Marks
A die is tossed thrice. Find the probability of getting an odd number at least once.
AnswerProbability of getting an odd number in a single throw of a die $=\frac{3}{6}=\frac{1}{2}$Similarly, probability of getting an even number $\frac{3}{6}=\frac{1}{2}$
Probability of getting an even number three times $=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}$
Therefore, probability of getting an odd number at least once
= 1 - Probability of getting an odd number in none of the throws
= 1 - Probability of getting an even number thrice
$=1-\frac{1}{8}$
$=\frac{7}{8}$
View full question & answer→Question 122 Marks
Which of the following distributions of a random variable X are the probability distributions?
| X: |
0 |
1 |
2 |
3 |
| P(X): |
0.3 |
0.2 |
0.4 |
0.1 |
AnswerP(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.3 + 0.2 + 0.4 + 0.1
= 1
It is the probability distribution of random variable X.
View full question & answer→Question 132 Marks
If $\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11}$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{11},$ find
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
AnswerGiven,
$\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11},\text{P}(\text{A}\cup\text{B})=\frac{7}{11}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{\frac{4}{11}}{\frac{6}{11}}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{2}{3}$
View full question & answer→Question 142 Marks
6 boys and 6 girls sit in a row at random. Find the probability that all the girls sit together.
AnswerHere, 6 bhoys and 6 girls can be arranged in a line in 12! ways.
Total possible outcomes = 12!
Consider 6 girls as a single element X.
Now, 6 boys and X can be arranged in aline in 7! ways and girls can be arranged in 6! ways among them.
P(all girls are together) $=\frac{7!\times6!}{12!}$
$=\frac{7\times6\times5\times4\times3\times2\times1\times6\times5\times4\times3\times2\times1}{12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}$
$=\frac{1}{11\times12}$
$=\frac{1}{132}$
View full question & answer→Question 152 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\text{A}\cup\text{B})$
AnswerGiven that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\text{A}\cup\text{B})=\text{P(A)} + \text{P(B)} - \text{P}(\text{A}\cap\text{B})$
$=0.3+0.6-0.18$
$\text{P}(\text{A}\cup\text{B})=0.72$
View full question & answer→Question 162 Marks
How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?
AnswerLet n be number of times a fair coin is tossed.
$\text{Now}\ \text{P}=\frac{1}{2},\ \text{q}=1-\text{p}=1-\frac{1}{2}=\frac{1}{2}\ (\text{given})$
P(at least one head) >90%
$\Rightarrow\ 1-\text{P}(0)>\frac{90}{100}\ \Rightarrow\ 1-\ ^\text{n}\text{C}_0\ \text{q}^\text{n}=\frac{9}{10}$
$\Rightarrow\ 1-1\times\bigg(\frac{1}{2}\bigg)^\text{n}>\frac{9}{10}\ \Rightarrow\ 1-\frac{9}{10}>\frac{1}{2^\text{n}}$
$\Rightarrow\ \frac{1}{10}>\frac{1}{2^\text{n}}\ \Rightarrow\ 2^\text{n}>10$
$\therefore$ least valus of n is 4
$\therefore$ minimum number of tosses is 4.
View full question & answer→Question 172 Marks
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
one of them is black and other is red.
AnswerTotal number of balls = 18
Number of red balls = 8
Number of black balls = 10
Probability of getting first ball as red $=\frac{8}{18}=\frac{4}{9}$
The ball is replaced after the first draw.
Probability of getting second ball as black $=\frac{10}{18}=\frac{5}{9}$
Therefore, probability of getting first ball as black and second ball as red $=\frac{4}{9}\times\frac{5}{9}=\frac{20}{81}$
Therefore, probability that one of them is black and other is red
= Probability of getting first ball black and second as red + Probability of getting first ball red and second ball black
$=\frac{20}{81}+\frac{20}{81}$
$=\frac{40}{81}$
View full question & answer→Question 182 Marks
Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
AnswerThere are 26 black cards in a deck of 52 cards.Let P(A) be the probability of getting a black card in the first draw.
$\therefore\ \text{P}(\text{A})=\frac{26}{52}=\frac{1}{2}$
Let P(B) be the probability of getting a black card on the second draw.Since the card is not replaced,
$ \therefore\ \text{P}(\text{B})=\frac{25}{51}$
Thus, probability of getting both the cards black $=\frac{1}{2}\times\frac{25}{51}=\frac{25}{102}$
View full question & answer→Question 192 Marks
Write the probability that a number selected at random from the set of first 100 natural numbers is a cube.
AnswerNumber of cubes in first 100 natural numbers = 1,8,27,64
So, there are 4 cubes in first 100 natural numbers. Pgetting a cube from a set of first 100 natural numbers $=\frac{4}{100}=\frac{1}{25}$
View full question & answer→Question 202 Marks
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that.
One of them is black and other is red.
AnswerGiven,
Box contains 10 black and 8 red balls.
Two balls are drawn with replacement.
P (one of them red and other black)
$=\text{P}\big((\text{B}\cap\text{R})\cup(\text{R}\cap\text{B})\big)$
$=\text{P}(\text{B}\cap\text{R})+\text{P}(\text{R}\cap\text{B})$
$=\text{P(B) }\text{P(R)}+\text{P(R) }\text{P(B)}$
$=\frac{10}{18}\times\frac{8}{18}+\frac{8}{18}\times\frac{10}{18}$
$=\frac{20+20}{81}$
$=\frac{40}{81}$
Required probability $=\frac{40}{81}$
View full question & answer→Question 212 Marks
A factory produces bulbs. The probability that one bulb is defective is $\frac{1}{50}$ and they are packed in boxes of 10. From a single box, find the probability that.
none of the bulbs is defective.
AnswerLet getting a defective bulb from a single box is a success.
We have,
p = probability of getting a defective bulb $=\frac{1}{50}$
Also, $\text{q}=1-\text{p}=1-\frac{1}{50}=\frac{49}{50}$
Let X denote the number of success in a sample of 10 trials. Then,
X follows binomial distribution with parameters n = 10 and $\text{p}=\frac{1}{50}$
$\therefore\text{P(X = r})=\text{ }^{10}\text{C}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(10-\text{r})}=\text{ }^{10}\text{C}_{\text{r}}\big(\frac{1}{50}\big)^{\text{r}}\big(\frac{49}{50}\big)^{(10-\text{r})}=\frac{\text{ }^{10}\text{C}_{\text{r}}49^{(10-\text{r})}}{50^{10}},$ where r = 0, 1, 2, 3, .....,10
Now,
Required probility = P(none of the bulb is defective)
$=\text{P(X}=0)$
$=\frac{\text{ }^{10}\text{C}_049^{10}}{50^{10}}$
$=\frac{49^{10}}{50^{10}}$
$=\big(\frac{49}{50}\big)^{10}$
View full question & answer→Question 222 Marks
Find the probability distribution of the number of sixes in three tosses of a die.
Answer
| $\text{X}$ |
$\text{P(X)}$ |
| $0$ |
$\text{}^3\text{c}_0\big(\frac{1}{6}\big)^0\big(\frac{5}{6}\big)^{3-0}=\big(\frac{5}{6}\big)^3=\frac{125}{216}$ |
| $1$ |
$\text{ }^3\text{c}_1\big(\frac{1}{6}\big)^1\big(\frac{5}{6}\big)^{3-1}=3\big(\frac{1}{6}\big)\big(\frac{1}{6}\big)^2=\frac{25}{72}$ |
| $2$ |
$\text{ }^3\text{c}_2\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^{3-2}=3\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)=\frac{5}{72}$ |
| $3$ |
$\text{ }^3\text{c}_3\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)^{3-3}=\big(\frac{1}{6}\big)^3=\frac{1}{216}$ |
View full question & answer→Question 232 Marks
An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be atleast 4 successes.
AnswerHere = 6
Now, p + q = 1 and p = 2q
$\therefore\ 2\ \text{q}+\text{q}=1\ \Rightarrow\ \text{q}=\frac{1}{3},\ \therefore\ \text{p}=\frac{2}{3}.$
$\therefore$ P(at least 4 successes in 6 trials) $=\text{P}(\text{X}\leq4)=\text{P}(4)+\text{P}(5)+\text{P}(6)$
$=\ ^6\text{C}_4\ \text{q}^2\ \text{p}^4+\ ^6\text{C}_5\ \text{q}\ \text{p}^4+\ ^6\text{C}_6\ \text{p}^6$
$=\text{p}^4\big[\ ^6\text{C}_2\ \text{q}^2+\ ^6\text{C}_1\ \text{p}\ \text{q}+\ ^6\text{C}_0\ \text{p}^2\big]$
$=\bigg(\frac{2}{3}\bigg)^4\bigg\{\frac{6\times5}{1\times2}\bigg(\frac{1}{3}\bigg)^2+\frac{6}{1}.\frac{2}{3}.\frac{1}{3}+1\times\bigg(\frac{2}{3}\bigg)^2\bigg\}$
$=\frac{16}{81}\bigg\{\frac{15}{9}+\frac{12}{9}+\frac{4}{9}\bigg\}=\frac{16}{81}\times\frac{31}{9}=\frac{496}{729}.$
View full question & answer→Question 242 Marks
Suppose X has a binomial distribution with n = 6 and p = $\frac{1}{2}.$ Show that X = 3 is the most likely outcome.
Answerwe have $\text{ n}=6$ and $\text{p}=\frac{1}{2}$
$\therefore\text{q}=1-\text{p}=\frac{1}{2}$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^6\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{6-\text{r}},\text{r}=0,1,2,3,4,5,6$
$=\text{ }^6\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{6}$
$\text{P(X = r})=\frac{\text{ }^6\text{C}_{\text{r}}}{2^6}$
By substituting r = 0, 1, 2, 3, 4, 5 and 6, we get the following
distribution for X.
| $\text{x}$ |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
| $\text{P(X)}$ |
$\frac{1}{64}$ |
$\frac{6}{64}$ |
$\frac{15}{64}$ |
$\frac{20}{64}$ |
$\frac{15}{64}$ |
$\frac{6}{64}$ |
$\frac{1}{64}$ |
Comparing the probabilites, we get that X = 3 is the most likely outcome View full question & answer→Question 252 Marks
Two cards are drawn without replacement from a pack of 52 cards. Find the probability that.
Both are kings.
AnswerIn a deck of 52 cards. there are 4 kings. Two cards are drawn without replacement,
A = First card is king
B = Second card is king
P (Both drawn cards are king)
$=\text{P}(\text{A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{4}{52}\times\frac{3}{51}$
$=\frac{1}{221}$
Required Probability $=\frac{1}{221}$
View full question & answer→Question 262 Marks
If A and B are two events write the expression for the probability of occurrence of exactly one of two events.
AnswerP(exaxtly one of 2 events) $=\text{P}(\text{A}\cup\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\text{P(A)}+\text{P(B)}-2\text{P}(\text{A}\cap\text{B})$
View full question & answer→Question 272 Marks
An unbiased die with face marked 1, 2, 3, 4, 5, 6 is rolled four times. Out of 4 face values obtained, find the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5.
AnswerDie is rolled 4 times
n(S) = Number of elements in samplw space
n(S) = 6 × 6 × 6 × 6
Given,
Number obtained on face are not less than 2 and greater than 5.
E = Obtaining 2, 3, 4, 5 on die in four throw
n(E) = 4 × 4 × 4 × 4
$\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}$
$=\frac{4\times4\times4\times4}{6\times6\times6\times6}$
$=\frac{16}{81}$
View full question & answer→Question 282 Marks
The random variable X has a probability distribution P(X) of the following form, where k is some number :
$ \text{P}(\text{X}) = \begin{cases} \text{k, }\overline{\text{if}\ \text{x}=0} \\ \overline{ 2\text{k, }\text{if}\ \text{x}=1}\\3\text{k, }\text{if}\ \text{x}=2\\0,\ \text{otherwise} \end{cases}$
- Determine the value of k.
- Find $P(X < 2), P(X ≤ 2), P(X ≥ 2)$.
AnswerProbability distribution:
| $x_i$ |
0 |
1 |
2 |
| $P(x_i)$ |
k |
2k |
3k |
- P(X = 0) + P(X = 1) + P(X = 2) = 1
⇒ k + 2k + 3k = 1 ⇒ 6k = 1
$\Rightarrow\ \text{k}=\frac{1}{6}$
- P(X < 2) = P(X = 0) + P(X = 1)
$=\text{k}+2\text{k}=3\text{k}=3\times\frac{1}{6}=\frac{1}{2}$
P(X ≥ 2) = P(X = 2) = 3k $=3\times\frac{1}{6}=\frac{1}{2}$ View full question & answer→Question 292 Marks
A bag contains 4 red and 5 black balls, a second bag contains 3 red and 7 black balls. One ball is drawn at random from each bag, find the probability that the,
Balls are of the same colour.
AnswerGiven,
Bag (1) contains 4 red and 6 black balls.
Bag (2) contains 3 red and 7 black balls
One ball is drawn ar random from each bag.
P (Balls are of the same colour)
$=\text{P}\big((\text{B}_1\cap\text{B}_2)\cup(\text{R}_1\cap\text{R}_2)\big)$
$=\text{P}(\text{B}_1\cap\text{B}_2)+\text{P}(\text{R}_1\cap\text{R}_2)$
$=\text{P}(\text{B}_1)\text{P}(\text{B}_2)+\text{P}(\text{R}_1)\text{P}(\text{R}_2)$
$=\frac{5}{9}\times\frac{7}{10}+\frac{4}{9}\times\frac{3}{10}$
$=\frac{35}{90}+\frac{12}{90}$
$=\frac{47}{90}$
Required probability $=\frac{47}{90}$
View full question & answer→Question 302 Marks
If two events A and B are such that $\text{P}(\overline{\text{A}})=0.3,\text{P(B)}=0.4$ and $\text{P}(\text{A}\cap\overline{\text{B}})=0.5$ find $\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big).$
AnswerAccording to Baye's Theorem
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big)=\frac{\text{P}(\text{B}\cap(\overline{\text{A}}\cap\overline{\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\text{B}(\overline{\text{A}\cap\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{B}}(\overline{\text{A}\cap\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{B}}\cup(\text{A}\cup\text{B}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
Now $\overline{\text{B}}\cap\text{B}=\cup=\phi$
So, $\text{P}(\overline{\text{B}}\cup(\text{A}\cup\text{B}))=\phi$
$\therefore\ \text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big)=0$
View full question & answer→Question 312 Marks
When three dice are thrown, write the probability of getting 4 or 5 on each of the dice simultaneously.
AnswerThere dice are thrown
Given, 4 or 5 on each of the dice simultaneously
= {(4, 4, 4), (4, 4, 5), (4, 5, 4), (4, 5, 5), (5, 4, 4), (5, 4, 5), (5, 5, 4), (5, 5, 5)}
n(E) = 8
n(S) = 216
$\text{P(E)}=\frac{8}{216}$
$=\frac{1}{27}$
Required probability $=\frac{1}{27}$
View full question & answer→Question 322 Marks
A random variable X has the following probability distribution:
| Values of X: |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
| P(X) |
a |
3a |
5a |
7a |
9a |
11a |
13a |
15a |
17a |
Determine:
The Value of a.
AnswerHere,
| Values of X: |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
| P(X) |
a |
3a |
5a |
7a |
9a |
11a |
13a |
15a |
17a |
Since $\sum\text{P}(\text{X})=1$
P(0) + P(1) + P(0) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) = 1
⇒ a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1
⇒ 81a = 1
$\Rightarrow\text{a}=\frac{1}{81}$ View full question & answer→Question 332 Marks
One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
E: ‘the card drawn is a king or queen’
F: ‘the card drawn is a queen or jack’
AnswerIn a deck of 52 cards, 4 cards are kings, 4 cards are queens, and 4 cards are jacks.
$\therefore$ P(E) = P(the card drawn is a king or a queen) $=\frac{8}{52}=\frac{2}{13}$
$\therefore$ P(F) = P(the card drawn is a queen or a jack) $=\frac{8}{52}=\frac{2}{13}$
There are 4 cards which are king or queen and queen or jack.
$\therefore$ P(EF) = P(the card drawn is a king or a queen, or queen or a jack)
$=\frac{4}{52}=\frac{1}{13}$
$\text{P}(\text{E})\times\text{P}(\text{F})=\frac{2}{13}\cdot\frac{2}{13}=\frac{4}{169}\neq\frac{1}{13}$
$\Rightarrow\text{P}(\text{E})\cdot\text{P}(\text{F})\neq\text{P}(\text{EF})$
Therefore, the given events E and F are not independent.
View full question & answer→Question 342 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\text{A}\cap\text{B})$
AnswerGiven that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\text{A}\cap\text{B})=\text{P(A) }\text{P(B)}$
[Since, A and B are indenpendent events]
$=0.3\times0.6$
$\text{P}(\text{A}\cap\text{B})=0.18$
View full question & answer→Question 352 Marks
If A and B are two independent events such that P(A) = 0.3 and $=0.8\text{P}(\text{A}\cap\overline{\text{B}})$ Find P(B).
AnswerA and B are two independent events
$\therefore\ \text{P}(\text{A}\cap\overline{\text{B}})=\text{P(A)}+\text{P}(\overline{\text{B}})-\text{P}(\text{A}\cap\overline{\text{B}})$
$\Rightarrow\ 0.8=0.3+[1-\text{P(B)}]-\text{P(A)}\text{P}(\overline{\text{B}})$
$\Rightarrow 0.5=1-\text{P(B)}-0.3[1-\text{P(B)}]$
$\Rightarrow 0.5=1-\text{P(B)}=0.3+0.3\text{P(B)}$
$\Rightarrow\ 0.5= 0.7-\text{P(B)}[1-0.3]$
$\Rightarrow\ 0.7\text{P(B)} = 0.2$
$\Rightarrow\ \text{P(B)}=\frac{0.2}{0.7}=\frac{2}{7}$
View full question & answer→Question 362 Marks
Which of the following distributions of a random variable X are the probability distributions?
|
X:
|
3
|
2
|
1
|
0
|
-1
|
|
P(X):
|
0.3
|
0.2
|
0.4
|
0.1
|
0.05
|
AnswerP(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) + P(X = -1)
= 0.3 + 0.2 + 0.4 + 0.1 + 0.05
= 1.05 > 1
It is not the probability distribution of random variable X.
View full question & answer→Question 372 Marks
Find the mean of the following probability distribution:
| $\text{X}=\text{x}_\text{i}:$ |
$1$ |
$2$ |
$3$ |
| $\text{P}(\text{X}=\text{x}_\text{i}):$ |
$\frac{1}{4}$ |
$\frac{1}{8}$ |
$\frac{5}{8}$ |
Answer
| $\text{x}_\text{i}$ |
$\text{p}_\text{i}$ |
$\text{p}_\text{i}\text{x}_\text{i}$ |
| $1$ |
$\frac{1}{4}$ |
$\frac{1}{4}$ |
| $2$ |
$\frac{1}{8}$ |
$\frac{2}{8}$ |
| $3$ |
$\frac{1}{8}$ |
$\frac{15}{8}$ |
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{1}{4}+\frac{2}{8}+\frac{15}{8}=\frac{2+2+15}{8}=\frac{19}{8}$ View full question & answer→Question 382 Marks
Determine P(E|F) in Exercises.
Mother, father and son line up at random for a family picture.
E : Son on one end, F : Father in middle.
Answer$\text{S}=(\text{MFS, MSF, SFM, SMF, FMS, FSM})\ \ \ \ \Rightarrow\ \ \ \ \text{n}(\text{S})=6$
E : Son on one end
$\text{E}=(\text {MFS, SFM, SMF, FMS})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \text{n}(\text{E})=4$
F : Father in middle
$\text{F}=(\text {MFS, SFM})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \text{n}(\text{F})=2$
$\text{P}\left (\text{F}\right)=\frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{2}{6}=\frac{1}{3}$
$\therefore\ \ \ \ \text{E}\cap\text{F}=\left(\text{MFS, SFM}\right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \text{n}\left(\text {E}\cap\text{F}\right)=2$
$\therefore\ \ \ \ \text{P}\left(\text{E}\cap\text{F}\right)=\frac{\text{n}\left(\text{E}\ \cap\ \text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{2} {6}=\frac{1}{3}$
$\text{And}\ \ \ \ \text{P}\left(\text{E}|\text{F}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{\frac {1}{3}}{\frac{1}{3}}=1$
View full question & answer→Question 392 Marks
A bag contains $4$ red and $4$ black balls, another bag contains $2$ red and $6$ black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
AnswerLet A be the event hat ball drawn is red and let $E_1$ and $E_2$ be the events that the ball drawn is from the first bag and second bag respectively.$\text{P}(\text{E}_1)=\frac{1}{2},\ \text{P}(\text{E}_2)=\frac{1}{2},$
$\text{P}(\text{A}|\text{E}_1)=\text{P}(\text{drawing a red ball from bag I})=\frac{4}{8}=\frac{1}{2}$
$\text{P}(\text{A}|\text{E}_2)=\text{P}(\text{drawing a red ball from bag II})=\frac{2}{8}=\frac{1}{4}$
View full question & answer→Question 402 Marks
Two cards are drawn without replacement from a pack of 52 cards. Find the probability that.
The first is a heart and second is red.
AnswerThere are 13 heart and 26 red cards
Hearts are also red.
A = first card is heart.
B = second card is red.
P (First card is heart and second is red)
$=\text{P(A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{13}{52}\times\frac{25}{51}$
$=\frac{25}{204}$
Required probability $=\frac{25}{204}$
View full question & answer→Question 412 Marks
Let A and B be two independent events such that $P(A) = p_1$ and $P(B) = p_2$. Describe in words the events whose probabilities are:
$(1 - p_1)p_2$.
AnswerAs, $(1 - p_1)p_2$
$=[1-\text{P(A)}]\times\text{P(B)}=\text{P}(\overline{\text{A}})\times\text{P(B)}$
And, A and B are independent events.
i.e., $\text{P}(\overline{\text{A}})\times\text{P}(\text{B})=\text{P}(\text{A}\cap\text{B})$
So, $\text{P}(\text{A}\cap\text{B})=\text{p}_1\text{p}_2$
Hence, $(1 - p_1)p_2$ = P (A does not occur, but B occurs).
View full question & answer→Question 422 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\text{A}\cap\overline{\text{B}})$
AnswerGiven that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\text{A}\cap\overline{\text{B}})=\text{P(A)}-\text{P}(\text{A}\cap\text{B})$
$=0.3 - 0.18$
$\text{P}(\text{A}\cap\overline{\text{B}})=0.12$
View full question & answer→Question 432 Marks
An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting,
2 red balls.
AnswerAn urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement.
P(Getting 2 red balls)
$=\text{P}(\text{R}_1)\times\text{P}(\text{R}_2)$
$=\frac{7}{11}\times\frac{7}{11}$
$=\frac{49}{121}$
Required probability $=\frac{49}{121}$
View full question & answer→Question 442 Marks
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs.
at least one will fuse after 150 days of use.
AnswerLet X be the number of bulbs that fuse after 150 days.
X follows a binomial distribution with n = 5, p = 0.05 and q = 0.95
Or $\text{p}=\frac{1}{20}$ and $\text{q}=\frac{19}{20}$
$\text{P}(\text{X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{20}\big)^{\text{r}}\big(\frac{19}{20}\big)^{5-\text{r}}$
Probability that that at lesast one will fuse
$=\text{P}(\text{X}=1)+\text{P}(\text{X}=2)+\text{P}(\text{X}=3)+\text{P}(\text{X}=4)+\text{P}(\text{X}=5)$
$=1-\text{P}(\text{X}=0)$
$=1-\Big[\text{ }^5\text{C}_0\big(\frac{1}{20}\big)^0\big(\frac{19}{20}\big)^{5-0}\Big]$
$=1-\Big[\big(\frac{19}{20}\big)^5\Big]$
View full question & answer→Question 452 Marks
A discrete random variable $X$ has the probability distribution given below:
| $X:$ |
$0.5$ |
$1$ |
$1.5$ |
$2$ |
| $P(X):$ |
$k$ |
$k^2$ |
$2k^2$ |
$k$ |
Find the value of $k.$ AnswerWe know that,
$P(0.5) + P(1) + P(1.5) + P(2) = 1$
$k + k^2+ 2k^2 + k = 1$
$3k^2 + 2k - 1 = 0$
$3k^2 + 3k - k - 1 = 0$
$(3k - 1)(k + 1) = 0$
$\text{k}=\frac{1}{3}$ or $\text{k}=-1$
We know that $0\leq\text{P}(\text{X})\leq1$
$\therefore\ \text{k}=\frac{1}{3}$
View full question & answer→Question 462 Marks
If X denotesthe number on the uper face of a cubical die when it is thrown, find the mean of X.
AnswerA cubical die can show 1, 2, 3, 4, 5 or 6 on its face
| $\text{x}_\text{i}$ |
$\text{p}_\text{i}$ |
$\text{p}_\text{i}\text{x}_\text{i}$ |
| $1$ |
$\frac{1}{6}$ |
$\frac{1}{6}$ |
| $2$ |
$\frac{1}{6}$ |
$\frac{2}{6}$ |
| $3$ |
$\frac{1}{6}$ |
$\frac{3}{6}$ |
| $4$ |
$\frac{1}{6}$ |
$\frac{4}{6}$ |
| $5$ |
$\frac{1}{6}$ |
$\frac{5}{6}$ |
| $6$ |
$\frac{1}{6}$ |
$\frac{6}{6}$ |
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{1}{6}+\frac{2}{6}+\frac{3}{6}+\frac{4}{6}+\frac{5}{6}+\frac{6}{6}=\frac{21}{6}=3.5$ View full question & answer→Question 472 Marks
Let E and F be events with $\text{P}(\text{E})=\frac{3}{5},\ \text{P}(\text{F})=\frac{3}{10}\ \text{and}\ \text{P}(\text{E}\cap\text{F})=\frac{1}{5}.$ Are E and F independent?
AnswerIt is given that $\text{P}(\text{E})=\frac{3}{5},\ \text{P}(\text{F})=\frac{3}{10},\ \text{and}\ \text{P}(\text{EF})=\text{P}(\text{E}\cap\text{F})=\frac{1}{5}$$\text{P}(\text{E}).\text{P}(\text{F})=\frac{3}{5}\cdot\frac{3}{10}=\frac{9}{50}\neq\frac{1}{5}$
$\Rightarrow\text{P}(\text{E}).\text{P}(\text{F})\neq\text{P}(\text{EF})$
Therefore, E and F are not independent.
View full question & answer→Question 482 Marks
A factory produces bulbs. The probability that one bulb is defective is $\frac{1}{50}$ and they are packed in boxes of 10. From a single box, find the probability that.
exactly two bulbs are defective.
AnswerLet getting a defective bulb from a single box is a success.
We have,
p = probability of getting a defective bulb $=\frac{1}{50}$
Also, $\text{q}=1-\text{p}=1-\frac{1}{50}=\frac{49}{50}$
Let X denote the number of success in a sample of 10 trials. Then,
X follows binomial distribution with parameters n = 10 and $\text{p}=\frac{1}{50}$
$\therefore\text{P(X = r})=\text{ }^{10}\text{C}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(10-\text{r})}=\text{ }^{10}\text{C}_{\text{r}}\big(\frac{1}{50}\big)^{\text{r}}\big(\frac{49}{50}\big)^{(10-\text{r})}=\frac{\text{ }^{10}\text{C}_{\text{r}}49^{(10-\text{r})}}{50^{10}},$ where r = 0, 1, 2, 3, .....,10
Now,
Required probability = P(exactly two bulbs are defective)
$=\text{P(X}=2)$
$=\frac{\text{ }^{10}\text{C}_249^8}{50^{10}}$
$=\frac{45\times49^8}{50^{10}}$
View full question & answer→Question 492 Marks
In a binomial distribution, if n = 20 and q = 0.75, then write its mean.
Answer$\text{n}=20,\text{q}=0.75$
$\Rightarrow\text{p}=1-\text{q}=0.25$
$\text{Mean = np}=20(0.25)=5$
Thus, $\text{Mean = 5}$
View full question & answer→Question 502 Marks
Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among 100 students, what is the probability that:
You both enter the different sections?
AnswerP (Both enter different section)
= 1 - P(Both enter same section)
$=1-\frac{17}{33}$
$=\frac{16}{33}$
P(Both eneter different section) $=\frac{16}{33}$
View full question & answer→Question 512 Marks
A and B are two events such that P (A) ≠ 0. Find P(B|A), if
- A is a subset of B
- $\text{A}\cap\text{B}=\phi$
AnswerA and B are two events such that P (A) ≠ 0.
To find: P(B|A)
- $\text{A is a subset of B}\ \Rightarrow\ \text{A}\subset\text{B}$
$\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})$
$\therefore\ \text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\frac{\text{P}(\text{A})}{\text{P}(\text{A})}=1$
- $\text{A}\cap\text{B}=\phi$
$\therefore\ \text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\frac{0}{\text{P}(\text{A})}=0$ View full question & answer→Question 522 Marks
The probability distribution function oif a random variable X is given by
| $X_i$ |
0 |
1 |
2 |
| $P_i$ |
$3c^3$ |
$4c - 10c^2$ |
$5c - 1$ |
Where $c > 0$
Find: $P(X < 2)$. Answer$\text{P}(\text{X}<2)=\text{P}(0)+\text{P}(1)$
$=3\text{c}^3+4\text{c}-10\text{c}^2$
$=3\Big(\frac{1}{3}\Big)^3+4\Big(\frac{1}{3}\Big)-10\Big(\frac{1}{3}\Big)^2$
$=\frac{3}{27}+\frac{4}{3}-\frac{10}{9}$
$=\frac{1}{9}+\frac{4}{3}-\frac{10}{9}$
$=\frac{3}{9}$
$\therefore\ \text{P}(\text{x}<2)=\frac{1}{3}$
View full question & answer→Question 532 Marks
Determine P(E|F) in Exercises. A die is thrown three times,E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses.
AnswerSince a dice has six faces. Therefore n (S) = 6 × 6 × 6 = 216
E = (1, 2, 3, 4, 5, 6) × (1, 2, 3, 4, 5, 6) × (4)
F = (6) × (5) × (1, 2, 3, 4, 5, 6) $\Rightarrow$ n(F) = 1 × 1 × 6 = 6
$\text{P}\left(\text{F}\right)=\frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{6}{216}$
$\therefore\ \ \ \ \ \ \ \text{E}\cap\text{F}=\left(6, 5, 4\right)\ \ \ \ \ \Rightarrow\ \ \ \text{n}\left(\text{E}\cap\text{F}\right)=1$
$\therefore\ \ \ \ \ \ \ \text{P}\left(\text{E}\cap\text{F}\right)=\frac{\text{n}\left(\text{E}\cap\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{1}{216}$
$\text{And}\ \ \ \ \text{P}\left(\text{E}|\text{F}\right)=\frac{\text{P}\left(\text{E}\cap\text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{216}$
View full question & answer→Question 542 Marks
A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its cooler is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be,
Of the same colour.
AnswerBag contains 3 blue, 5 red marble. One marble is drawn, its colour noted and replaced then again a marble drawn and its colour is noted.
P (Of the same colour)
$=\text{P}\big((\text{R}_1\cap\text{R}_2)\cup(\text{B}_1\cap\text{B}_2)\big)$
$=\text{P}(\text{R}_1)\times\text{P}(\text{R}_2)+\text{P}(\text{B}_1)\times\text{P}(\text{B}_2)$
$=\frac{5}{8}\times\frac{5}{8}+\frac{3}{8}\times\frac{3}{8}$
$=\frac{25+9}{64}$
$=\frac{34}{64}$
$=\frac{17}{32}$
View full question & answer→Question 552 Marks
An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting,
One red and one blue ball.
AnswerAn urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement.
P(Getting 2 blue balls)
$=\text{P}\big((\text{R}\cap\text{B})\cup(\text{B}\cap\text{R})\big)$
$=\text{P(R)}\times\text{P(B)}+\text{P(B)}\times\text{P(R)}$
$=\frac{7}{11}\times\frac{4}{11}+\frac{4}{11}\times\frac{7}{11}$
$=\frac{28+28}{121}$
$=\frac{56}{121}$
Required probability $=\frac{56}{121}$
View full question & answer→Question 562 Marks
If $\text{P(A)}=0.3,\text{P(B)}=0.6,\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.5,$ find $\text{P}(\text{A}\cup\text{B}).$
AnswerGiven,
$\text{P(A)}=0.3,\text{P(B)}=0.6,\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.5,$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$0.5=\frac{\text{P}(\text{A}\cap\text{B})}{0.3}$
$\text{P}(\text{A}\cap\text{B})=0.5\times0.3$
$\text{P}(\text{A}\cap\text{B})=0.15$
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.3+0.6+0.15$
$=0.75$
$\text{P}(\text{A}\cup\text{B})=0.75$
View full question & answer→Question 572 Marks
The mean and variance of a binomial distribution are $\frac{4}{3}$ and $\frac{8}{9}$ respectively. Find P (X ≥ 1).
Answer$=1-\big(\frac{2}{3}\big)^{4}$
$=1-\frac{16}{81}$
$=\frac{65}{81}$
$\text{P(X}\geq1)=\frac{65}{81}$
View full question & answer→Question 582 Marks
Three digit numbers are formed with the digits 0, 2, 4, 6 and 8. Write the probability of forming a three digit number with the same digits.
AnswerTotal 3-digit numbers that can be made out of 0, 2, 4, 5 and 8 = 4 × 5 × 5 (hundreds place cannot be filled with 0)
= 100
But 222, 444, 666 and 888 are four numbers, which have the same digits at all places.
P(3-digit number having same digits at all places) $=\frac{4}{100}$
$=\frac{1}{25}$
View full question & answer→Question 592 Marks
An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting,
2 Blue balls.
AnswerAn urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement.
P(Getting 2 blue balls)
$=\text{P}(\text{B}_1\cap\text{B}_2)$
$=\text{P}(\text{B}_1)\times\text{P}(\text{B}_2)$
$=\frac{4}{11}\times\frac{4}{11}$
$=\frac{16}{121}$
Required probability $=\frac{16}{121}$
View full question & answer→Question 602 Marks
An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable ?
AnswerThere two balls may be selected as BR, RB, BR, BB, where R represents red ball and B represents black ball.
Variable X has the value 0, 1, 2, i.e., there may be no black ball, may be one black ball or both the balls are black.
View full question & answer→Question 612 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\overline{\text{A}}\cap\text{B})$
AnswerGiven that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.6 - 0.18$
$\text{P}(\overline{\text{A}}\cap\text{B})=0.42$
View full question & answer→Question 622 Marks
A bag contains $4$ white balls and $2$ black balls. Another contains $3$ white balls and $5$ black balls. If one ball is drawn from each bag, find the probability that,
One is while and one is black.
AnswerBag A has 4 white balls and 2 black balls;
Bag B has 3 white balls and 5 black balls;
$P (A_W$ and $B_B$ or $A_B$ and $B_W)$
$= P(A_W) P(B_B) + P(A_B) P(B_W)$
$=\frac{4}{6}\times\frac{5}{8}+\frac{2}{6}\times\frac{3}{8}$
$=\frac{20}{48}+\frac{6}{48}$
$=\frac{26}{48}=\frac{13}{24}$
View full question & answer→Question 632 Marks
If the probability distribution of a random variable X is given below:
| $X = X_i$ |
1 |
2 |
3 |
4 |
| $P(X = X_i)$ |
c |
2c |
4c |
4c |
Write the value of $\text{P}(\text{X}\leq2)$ AnswerWe know that the sum of probabilities in a probability distribution is always 1.
Therefore,
$P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1$
$\Rightarrow c + 2c + 4c + 4c =1$
$\Rightarrow 11c = 1$
$\Rightarrow\text{c}=\frac{1}{11}$
Now,
$\text{P}(\text{X}\leq2)=\text{P}(\text{X}=1)+\text{P}(\text{X})=2=\frac{1}{10}+\frac{2}{10}=\frac{3}{10}$
View full question & answer→Question 642 Marks
A bag contains 6 black and 3 white balls. Another bag contains 5 black and 4 white balls. If one ball is drawn from each bag, find the probability that these two balls are of the same colour.
AnswerGiven:
Bag 1 = (3W + 6B) balls
Bag 2 = (5B + 4W) balls
P(balls of same colour are drawn) = P(both black) + P(both white)
$=\frac{6}{9}\times\frac{5}{9}+\frac{3}{9}\times\frac{4}{9}$
$=\frac{30}{81}+\frac{12}{81}$
$=\frac{42}{81}$
$=\frac{14}{27}$
View full question & answer→Question 652 Marks
A fair coin is tossed 8 times, find the probability of.
at least six heads.
AnswerProbability of getting atleast 6 heads
$=\text{P}(\text{X}\geq6)$
$=\text{P}(\text{X}=6)+\text{P}(\text{X}=7)+\text{P}(\text{X}=8)$
$=\text{ }^8\text{C}_6\big(\frac{1}{2}\big)^8+\text{ }^8\text{C}_7\big(\frac{1}{2}\big)^8+\text{ }^8\text{C}_8\big(\frac{1}{2}\big)^8$
$=(28+8+1)\times\frac{1}{256}$
$=\frac{37}{256}$
View full question & answer→Question 662 Marks
Three cards are drawn with replacement from a well shuffled pack of cards. Find the probability that the cards drawn are king, queen and jack.
Answer$\text{P (king)}=\frac{4}{52}$
$\text{P (queen)}=\frac{4}{52}$
$\text{P (jack)}=\frac{4}{52}$
These cards can be drawn in $^3P_3$ ways.
$\text{P (king, queen, and jack)}=\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}\times {^{3}}\text{P}_3$
$=\frac{3!}{2197}$
$=\frac{6}{2197}$
View full question & answer→Question 672 Marks
Six coins are tossed simultaneously. Find the probability of getting.
no heads.
AnswerLet p represents the probability of getting head in a toss of fair coin, so
$\text{p}=\frac{1}{2}$
$\text{q}=1-\frac{1}{2}$ [Since p + q = 1]
$\text{q}=\frac{1}{2}$
Let X denote the random variable representing the number heads in 6 tosses of coin. probability of getting r sixes in n tosses of a fair coin is given by,
$\text{P(X = r})=\text{ }^\text{n}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$3
$=\text{ }^6\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{6-\text{r}}\dots(1)$
P(getting no head) = P(X = 0)
$=\frac{\text{ }^6\text{C}_0}{2^6}$
$=\big(\frac{1}{2}\big)^6$
$=\frac{1}{64}$
View full question & answer→Question 682 Marks
If in a binomial distribution mean is 5 and variance is 4, write the number of trials.
Answer$\text{Mean}=5$ and $\text{Variance}=4$
$\Rightarrow\text{np}=5$ and $\text{npq}=4$
$\Rightarrow\text{q}=0.8$
$\Rightarrow\text{p}=1-\text{q}=0.2$
$\&\text{ np}=\text{n}(0.2)=5$ (given)
$\Rightarrow\text{n}=\frac{5}{0.2}=25$
View full question & answer→Question 692 Marks
A four digit number is formed using the digits 1, 2, 3, 5 with no repetitions. Write the probability that the number is divisible by 5.
AnswerTo be divisibel by 5 ones place sholud be 5
There are 3 place remaining which can be filled in 3! = 6 ways
So, 6 number can be formed out of 1, 2, 3 and 5, which are divisible by 5.
Total 4 - digit numbers = 4! = 24
P (4-digit number divisible by 5) $=\frac{6}{24}$
$=\frac{1}{4}$
View full question & answer→Question 702 Marks
Five cards are drawn successively with replacement from a well-shuffled pack of 52 cards. What is the probability that
all the five cards are spades?
AnswerLet P denote the probability of geting one spade out of a deck of 52 cards, So
$\text{p}=\frac{13}{52}$
$\text{p}=\frac{1}{4}$
$\text{q}=1-\frac{1}{4}$ [Since p + q = 1]
$\text{q}=\frac{3}{4}$
let X denote the radom variable of number of spades out of 5 cards. probability of getting r spades out of n cards is given by
$\text{P}(\text{x = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^5\text{c}_\text{r}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{5-\text{r}}\dots(1)$
Probability of getting all five spads
$\text{P}(\text{X}=5)$
$=\text{ }^5\text{c}_5\big(\frac{1}{4}\big)^5\big(\frac{3}{4}\big)^5-5$
$=\frac{1}{1024}$
Probability of getting 5 spades $=\frac{1}{1024}$
View full question & answer→Question 712 Marks
A bag contains $4$ white balls and $2$ black balls. Another contains $3$ white balls and $5$ black balls. If one ball is drawn from each bag, find the probability that,
Both are white.
AnswerBag A has 4 white balls and 2 black balls;
Bag B has 3 white balls and 5 black balls;
$P (A_W$ and $B_W) = P(A_W) P(B_W)$
$=\frac{4}{6}\times\frac{3}{8}=\frac{1}{4}$
View full question & answer→Question 722 Marks
Which of the following distributions of a random variable X are the probability distributions?
|
X:
|
0
|
1
|
2
|
|
P(X):
|
0.6
|
0.4
|
0.2
|
AnswerP(X = 0) + P(X = 1) + P(X = 1) + P(X = 2)
= 0.6 + 0.4 + 0.2
= 1.2 > 1
It is not the probability distribution of random variable X.
View full question & answer→Question 732 Marks
Which of the following distributions of a random variable X are the probability distributions?
| X: |
0 |
1 |
2 |
3 |
4 |
| P(X): |
0.1 |
0.5 |
0.2 |
0.1 |
0.1 |
AnswerP(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= 0.1 + 0.5 + 0.2 + 0.1 + 0.1
= 1
It is the probability distribution of random variable X.
View full question & answer→Question 742 Marks
Two cards are drawn from a well shuffled pack of 52 cards, one after another without replacement. Find the probability that one of these is red card and the other a black card?
AnswerP(One red and one black) = P(first red and second black) + P(first black and second red)
$=\frac{26}{5}\times\frac{26}{51}+\frac{26}{52}\times\frac{26}{51}$
[Without replacement]
$=\frac{13}{51}+\frac{13}{51}$
$=\frac{26}{51}$
View full question & answer→Question 752 Marks
A bag contains 2 white, 3 red and 4 blue balls. Two balls are drawn at random from the bag. If X denotes the number of white balls among the two balls drawn, describe the probability distribution of X.
Answer
| $\text{X}$ |
$\text{P(X)}$ |
| $0$ |
$\frac{7}{6}\times\frac{6}{8}=\frac{21}{36}$ |
| $1$ |
$\frac{7}{9}\times\frac{2}{8}\times2=\frac{14}{36}$ |
| $2$ |
$\frac{2}{9}\times\frac{1}{8}=\frac{1}{36}$ |
View full question & answer→Question 762 Marks
If on an average 9 ships out of 10 arrive safely at ports, find the mean and S.D. of the ships returning safely out of a total of 500 ships.
AnswerTotal number of ships (n) = 500
Let X denote the number of ships returning safely to the ports.
$\text{p}=\frac{9}{10}$ and $\text{q}=1-\text{p}=\frac{1}{10}$
$\text{Mean = np = 450}$ and $\text{Variance = npq = 45}$
$\text{Mean = 450}$
$\text{Standard deviation}=\sqrt{45}=6.71$
View full question & answer→Question 772 Marks
Six coins are tossed simultaneously. Find the probability of getting.
at least one head.
AnswerLet p represents the probability of getting head in a toss of fair coin, so
$\text{p}=\frac{1}{2}$
$\text{q}=1-\frac{1}{2}$ [Since p + q = 1]
$\text{q}=\frac{1}{2}$
Let X denote the random variable representing the number heads in 6 tosses of coin. probability of getting r sixes in n tosses of a fair coin is given by,
$\text{P(X = r})=\text{ }^\text{n}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$3
$=\text{ }^6\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{6-\text{r}}\dots(1)$
P(getting at least 1 head) $=\text{P(X}\geq1)$
$=1-\text{P(X}=0)$
$=1-\frac{1}{64}$
$=\frac{63}{64}$
View full question & answer→Question 782 Marks
A card is drawn from a well-shulffled deck of 52 cards. The outcome is noted, the card is replaced and the deck reshuffled. Another card is then drawn from the deck.
What is the probability that both the cards are of the same suit?
AnswerP(both the cards are of same suit) = P(both the cards are of diamond) + P(both the cards are of spade) + P(both the cards are of club) + P(both the cards are of hear)
$=\frac{13}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}$
$=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}$
$=\frac{4}{16}$
$=\frac{1}{4}$
View full question & answer→Question 792 Marks
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs.
none.
AnswerLet X be the number of bulbs fuse after 150 days.
X follows a binomial disribution with n = 5, p = 0.05 and q = 0.95
Or $\text{p}=\frac{1}{20}$ and $\text{q}=\frac{19}{20}$
$\text{P}(\text{X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{20}\big)^{\text{r}}\big(\frac{19}{20}\big)^{5-\text{r}}$
Probability (nine will fuse after 150 days of use) = P(X = 0)
$=\text{ }^5\text{C}_0\big(\frac{1}{20}\big)^0\big(\frac{19}{20}\big)^{5-0}$
$=\big(\frac{19}{20}\big)^5$
View full question & answer→Question 802 Marks
If $\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13},$ find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big).$
AnswerGiven:
$\text{P(A)}=\frac{7}{13}$
$\text{P(B)}=\frac{9}{13}$
$\text{P}(\text{A}\cap\text{B})=\frac{4}{13}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{4}{13}}{\frac{9}{13}}=\frac{4}{9}$
View full question & answer→Question 812 Marks
Find the expected number of boys in a family with 8 children, assuming the sex distribution to be equally probable.
AnswerHere, $\text{n}=8$
Let p be the probability of number of boys in the family.
$\text{p}=\frac{1}{2},\text{q}=\frac{1}{2}$
Expected number of boys = Mean
$\Rightarrow\text{np}=4$
View full question & answer→Question 822 Marks
In a competition A, B and C are participating. The probability that A wins is twice that of B, the probability that B wins is twice that of C. Find the probability that A losses.
AnswerLet P(A wins) = x
So, P(B wins) = 2x
P(A wins) = 2P(B wins)
= 2(2x)
P(A wins) = 4X
P(A wins) + P(B wins) + P(C wins) = 1
⇒ 4x + 2x+ x = 1
⇒ 7x = 1
$\Rightarrow\ \text{x}=\frac{1}{7}$
$\text{P(A wins)}=4\text{x}$
$=\frac{4}{7}$
$\text{P(A losses)}=1-\text{P(A wins)}$
$=1-\frac{4}{7}$
$=\frac{3}{7}$
Required probability $=\frac{3}{7}$
View full question & answer→Question 832 Marks
A random variable has the following probability distribution:
| $X = X_i$ |
1 |
2 |
3 |
4 |
| $P(X = X_i)$ |
k |
2k |
3k |
4k |
Write the value of $\text{P}(\text{X}\geq3).$ AnswerHere,
| $X = X_i$ |
1 |
2 |
3 |
4 |
| $P(X = X_i)$ |
k |
2k |
3k |
4k |
Since, $\sum\text{P}(\text{X})=1$ ⇒ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1 ⇒ k + 2k + 3k + 4k =1 ⇒ 10k = 1 $\Rightarrow\text{k}=\frac{1}{10}$ $\text{P}(\text{X}\geq3)$= P(X = 3) + P(X = 4)
= 3k + 4k
= 7k $=\frac{1}{10}$ $\text{P}(\text{X}\geq3)=\frac{7}{10}$ View full question & answer→Question 842 Marks
A ordinary cube has four plane faces, one face marked $2$ and another face marked $3$, find the probability of getting a total of $7$ in $5$ throws.
AnswerA cube has total 6 faces.
Total possible outcomes in 5 throws = $6 \times 6\times 6 \times 6 \times 6 = (6)^5$
The only way of getting 7 is by getting two 2s and one 3.
Total possible ways $=\frac{\text{P}^5_3}{2!}$
$=\frac{5\times4\times3\times2\times1}{2\times1\times2\times1}$
$=30$
Now,
P(getting 7 in 5 throws) $=\frac{30}{6^5}=\frac{5}{6^4}$
View full question & answer→Question 852 Marks
Suppose that a radio tube inserted into a certain type of set has probability $0.2$ of functioning more than $500$ hours. If we test $4$ tubes at random what is the probability that exactly three of these tubes function for more than $500$ hours$?$
AnswerLet $X$ denote the number of tubes that function for more than $500$ hours.
Then, $X$ follows a binomial distribution with $n = 4.$
Let $p$ be the probability that the tubes function more than $500$ hours
Here, $p = 0.2, q = 0.8$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^4\text{C}_{\text{r}}(0.2)^{\text{r}}(0.8)^{4-\text{r}},\text{r}=0,1,2,3,4$
Therefore, required probability $= P(X = 3)$
$= 4(0.2)^3(0.8)$
$= 0.0256$
View full question & answer→Question 862 Marks
A random variable X has the following probability distribution:
| Values of X: |
-2 |
-1 |
0 |
1 |
2 |
3 |
| P(X) |
0.1 |
k |
0.2 |
2k |
0.3 |
k |
Find the value of k.
AnswerHere,
| Values of X: |
-2 |
-1 |
0 |
1 |
2 |
3 |
| P(X) |
0.1 |
k |
0.2 |
2k |
0.3 |
k |
We know that,
P(-2) + P(-1) + P(0) + P(1) + P(2) + P(3) = 1
⇒ 0.1 + k + 0.2 + 2k + 0.3 + k = 1
⇒ 4k + 0.6 = 1
⇒ 4k = 1 - 0.6
⇒ 4k = 0.4
$\Rightarrow\text{k}=\frac{0.4}{4}$
$\Rightarrow\text{k}=\frac{1}{10}$
$\Rightarrow\text{k}=0.1$ View full question & answer→Question 872 Marks
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that.
First ball is black and second is red.
AnswerGiven,
Box contains 10 black and 8 red balls.
Two balls are drawn with replacement.
P (First ball is black and second is red)
$=\text{P}(\text{B}\cap\text{R})$
$=\text{P(B)}\text{ P(R)}$
$=\frac{10}{18}\times\frac{8}{18}$
$=\frac{20}{81}$
Required probability $=\frac{20}{81}$
View full question & answer→Question 882 Marks
A factory produces bulbs. The probability that one bulb is defective is $\frac{1}{50}$ and they are packed in boxes of 10. From a single box, find the probability that.
more than 8 bulbs work properly.
AnswerLet getting a defective bulb from a single box is a success.
We have,
p = probability of getting a defective bulb $=\frac{1}{50}$
Also, $\text{q}=1-\text{p}=1-\frac{1}{50}=\frac{49}{50}$
Let X denote the number of success in a sample of 10 trials. Then,
X follows binomial distribution with parameters n = 10 and $\text{p}=\frac{1}{50}$
$\therefore\text{P(X = r})=\text{ }^{10}\text{C}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(10-\text{r})}=\text{ }^{10}\text{C}_{\text{r}}\big(\frac{1}{50}\big)^{\text{r}}\big(\frac{49}{50}\big)^{(10-\text{r})}=\frac{\text{ }^{10}\text{C}_{\text{r}}49^{(10-\text{r})}}{50^{10}},$ where r = 0, 1, 2, 3, .....,10
Now,
Required probility = P(more than 8 bulbs work properly)
= P(atmost one bulb is defective)
$=\text{P(X}\leq0)$
$=\text{P(X}=0)+\text{P(X}=1)$
$=\frac{\text{ }^{10}\text{C}_049^{10}}{50^{10}}+\frac{\text{ }^{10}\text{C}_149^9}{}$
$=\frac{49^{10}}{50^{10}}+\frac{10\times49^{9}}{50^{10}}$
$=\frac{49^6}{50^{10}}(49+10)$
$=\frac{59(49^9)}{(50^{10})}$
View full question & answer→Question 892 Marks
Let A and B be two independent events such that $P(A) = p_1$ and $P(B) = p_2$. Describe in words the events whose probabilities are:
$p_1p_2$.
AnswerAs, $p_1p_2 = P(A) \times P(B)$
And, A and B are independent events.
i.e., $\text{P(A)}\times\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
So, $\text{P}(\text{A}\cap\text{B})=\text{P}_1\text{P}_2$
Hence, $p_1p_2$ = P(A and B occur).
View full question & answer→Question 902 Marks
Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that,
- The youngest is a girl.
- At least one is a girl?
AnswerLet a and g rapresent the boyy and the girl child respectively. if a family has two children, the samplw space will be
S = {(b, b), (b, g), (g, b), (g, g)}
Let a be the event that both children are girls.
$\therefore\ \text{A}=\big\{(\text{g},\text{g})\big\}$
- Let B be the event that the youngest child is a girl.
$\therefore\text{B}=\big[(\text{b},\text{g}),(\text{g},\text{g})\big]$
$\Rightarrow\ \text{A}\cap\text{B}=\big\{(\text{g},\text{g})\big\}$
$\therefore\text{P(B)}=\frac{2}{4}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{4}$
The conditional probability that both are girls, given that the youngest child is a girl, is given by p (A | B).
$\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$
Therefore, the required probability is $\frac{1}{2}$.
- Let C be the event that at least one child is a girl.
$\therefore\text{C}=\big\{(\text{b},\text{g}),(\text{g},\text{b}),(\text{g},\text{g})\big\}$
$\Rightarrow\text{A}\cap\text{C}=\big\{\text{g},\text{g}\big\}$
$\Rightarrow\text{P(C)}=\frac{3}{4}$
$\text{P}(\text{A}\cap\text{C})=\frac{1}{4}$
The conditional probability that both are girls, given that at least one child a girl, is given by P(A|C).
$\therefore \text{P}(\text{A}|\text{C})=\frac{\text{P}(\text{A}\cap\text{C})}{\text{P}(\text{C})}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$ View full question & answer→Question 912 Marks
The probability distribution of random variable X is given below:
|
$\text{X}$
|
$0$
|
$1$
|
$2$
|
$3$
|
|
$\text{P}(\text{X})$
|
$\text{k}$
|
$\frac{\text{k}}{2}$
|
$\frac{\text{k}}{4}$
|
$\frac{\text{k}}{8}$
|
Determine $\text{P}(\text{X}\leq2)$ and $\text{P}(\text{X}>2)$ Answer$\text{P}(\text{X}\leq2)$
$=\text{P}(0)+\text{P}(1)+\text{P}(2)$
$=\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}$
$=\frac{8}{15}+\frac{8}{30}+\frac{8}{60}$
$=\frac{14}{15}$
$\text{P}(\text{X}>2)=\text{P}(3)=\frac{\text{k}}{8}=\frac{1}{15}$
View full question & answer→Question 922 Marks
The probability distribution of random variable X is given below:
|
$\text{X}$
|
$0$
|
$1$
|
$2$
|
$3$
|
|
$\text{P}(\text{X})$
|
$\text{k}$
|
$\frac{\text{k}}{2}$
|
$\frac{\text{k}}{4}$
|
$\frac{\text{k}}{8}$
|
Determine the value of k. AnswerWe know that,
$\text{P}(0)+\text{P}(1)+\text{P}(2)+\text{P}(3)=1$
$\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}+\frac{\text{k}}{8}=1$
$15\text{k}=8$
$\text{k}=\frac{8}{15}$
View full question & answer→Question 932 Marks
X is taking up subjects - Mathematics, Physics and Chemistry in the examination. His probabilities of getting grade A in these subjects are 0.2, 0.3 and 0.5 respectively. Find the probability that he gets,
Grade A in no subject.
AnswerGiven,
Probability of getting A grade in mathematics (m) = 0.2
⇒ P(m) = 0.2
Probability of getting A grade in physics (p) = 0.3
⇒ P(p) = 0.3
Probability of getting A grade in chemistry (P) = 0.5
⇒ P(C) = 0.5
P(Getting A in no subject)
$=\text{P}(\overline{\text{m}}\cap\overline{\text{p}}\cap\overline{\text{c}}\big)$
$=\text{P}(\overline{\text{m}})+\text{P}(\overline{\text{p}})+\text{P}(\overline{\text{c}}\big)$
$=\big(1-\text{P(m)}\big)\big(1-\text{P(p)}\big)\big(1-\text{P(c)}\big)$
$=(1-0.2)(1-0.3)(1-0.5)$
$=(0.8)(0.7)(0.5)$
$=0.28$
Required probability = 0.28
View full question & answer→Question 942 Marks
Fatima and John appear in an interview for two vacancies for the same post. The probability of Fatima's selection is $\frac{1}{7}$ and that of John's selection is $\frac{1}{5}$. What is the probability that,
None of them will be selected?
AnswerGiven,
Probability of Fatima's (F) selection $=\frac{1}{7}$
$\text{P(F)}=\frac{1}{7}\Rightarrow\ \text{P}(\overline{\text{F}})=\frac{6}{7}$
Probability of John's (J) selection $=\frac{1}{5}$
$\text{P(F)}=\frac{1}{5}\Rightarrow\ \text{P}(\overline{\text{F}})=\frac{4}{5}$
P(None of them selected)
$=\text{P}(\overline{\text{F}}\cap\overline{\text{J}})$
$=\text{P}(\overline{\text{F}})+\text{P}(\overline{\text{J}})$
$=\frac{6}{7}\times\frac{4}{5}$
$=\frac{24}{35}$
Required probabilty $=\frac{24}{35}$
View full question & answer→Question 952 Marks
The probability distribution function oif a random variable X is given by
| $X_i$ |
0 |
1 |
2 |
| $P_i$ |
$3c^3$ |
$4c - 10c^2$ |
$5c - 1$ |
Where c > 0
Find: $\text{P}(1<\text{X}\leq2)$ Answer$\text{P}(1<\text{X}\leq2)$
$=\text{P}(\text{X}=2)$
$=5\text{c}-1$
$=\frac{5}{3}-1$
$=\frac{5-3}{3}$
$=\frac{2}{3}$
View full question & answer→Question 962 Marks
One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
E: ‘the card drawn is black’
F: ‘the card drawn is a king’
AnswerIn a deck of 52 cards, 26 cards are black and 4 cards are kings.
$\therefore$ P(E) = P(the card drawn is black) $=\frac{26}{52}=\frac{1}{2}$
$\therefore$ P(F) = P(the card drawn is a king) $=\frac{4}{52}=\frac{1}{13}$
In the pack of 52 cards, 2 cards are black as well as kings.
$\therefore$ P (EF) = P(the card drawn is a black king) $=\frac{2}{52}=\frac{1}{26}$
$\text{P}(\text{E})\times\text{P}(\text{F})=\frac{1}{2}\cdot\frac{1}{13}=\frac{1}{26}=\text{P}(\text{EF})$
Therefore, the given events E and F are independent.
View full question & answer→Question 972 Marks
X is taking up subjects - Mathematics, Physics and Chemistry in the examination. His probabilities of getting grade A in these subjects are 0.2, 0.3 and 0.5 respectively. Find the probability that he gets,
Grade A in all subjects
AnswerGiven,
Probability of getting A grade in mathematics (m) = 0.2
⇒ P(m) = 0.2
Probability of getting A grade in physics (p) = 0.3
⇒ P(p) = 0.3
Probability of getting A grade in chemistry (P) = 0.5
⇒ P(C) = 0.5
P(Getting A grade in all subjects)
$=\text{P}(\text{m}\cap\text{P}\cap\text{C})$
= P(m) + P(p) + P(c)
= 0.2 × 0.3 × 0.5
= 0.3
Required probability = 0.03
View full question & answer→Question 982 Marks
A husband and wife appear in an interview for two vacancies for the same post. The probability of husband's selection is $\frac{1}{7}$ and that of wife's selection is $\frac{1}{5}$. What is the probability that,
Both of them will be selected?
AnswerGiven,
Probability of Husband's (H) selection $=\frac{1}{7}$
$\text{P(H)}=\frac{1}{7}$
Probability of wife's (W) selection $=\frac{1}{5}$
$\text{P(W)}=\frac{1}{5}$
P(Both of them will be selelcted)
$=(\text{H}\cap\text{W})$
$=\text{P(H)}\text{ P(W)}$
$=\frac{1}{7}\times\frac{1}{5}$
$=\frac{1}{35}$
Required probability $=\frac{1}{35}$
View full question & answer→Question 992 Marks
The mean of a binomial distribution is $10$ and its standard deviation is $2$; write the value of q.
AnswerMean of the binomial distribution, i.e. $\text{np}=10$
$Variance = (Standard\ deviation)^2$, i.e. $\text{npq}=4$
$\therefore\text{q}=\frac{\text{Variance}}{\text{Mean}}=0.4$
View full question & answer→Question 1002 Marks
If A and B are independent events, then write expression for P(exactly one of A, B occurs).
AnswerAs, A and B are independent events.
So, $\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P(B)}\ .....\text{(i)}$
P(exactly one of A, B occurs) = P(only A) + P(only B)
$=\big[\text{P(A)}-\text{P}(\text{A}\cap\text{B})\big]+\big[\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$
$=\big[\text{P(A)}-\text{P}(\text{A})\times\text{P}(\text{B})\big]+\big[\text{P(B)}-\text{P}(\text{A})\times\text{P}(\text{B})\big]$ [Using (i)]
$=\text{P(A)}\times\big[1-\text{P(B)}\big]+\text{P(B)}\times\big[1-\text{P(A)}\big]$
$=\text{P(A)}\times\text{P}(\overline{\text{B}})+\text{P(B)}\times\text{P}(\overline{\text{A}})$
View full question & answer→Question 1012 Marks
Kamal and Monica appeared for an interview for two vacancies. The probability of Kamal's selection is $\frac{1}{3}$ and that of Monika's selection is $\frac{1}{5}$. Find the probability that.
At least one of them will be selected.
AnswerP (Kamal gets selected) $=\text{P(A)}=\frac{1}{3}$
P (Monica gets selected) $=\text{P(B)}=\frac{1}{5}$
P (At least one of them gets selected) $=\text{P}(\text{A}\cup\text{B})$
$=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})$
$=\text{P(A)}+\text{P(B)}-\text{P(A)}\times\text{P(B)}$
$=\frac{1}{3}+\frac{1}{5}-\frac{1}{3}\times\frac{1}{5}$
$=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}$
$=\frac{7}{15}$
View full question & answer→Question 1022 Marks
Three numbers are chosen from $1$ to $20$. Find the probability that they are consecutive.
Answer$S =$ There numbers are chosen from $1$ to $20$
$n(S) = {^{20}C_3}$
$E =$ Group of three consecutive numbers between $1$ and $20$
$n(E) = 18$
$\{(1, 2, 3,), (2, 3, 4), (3, 4, 5), (4, 5, 6), (5, 6, 7), (6, 7, 8), (7, 8, 9), (8, 9, 10), (9, 10, 11), (10, 11, 12), (11, 12, 13), (11, 12, 13), (12, 13, 14), (13, 14, 15), (14, 15, 16), (15, 16, 17), (16, 17, 18), (17, 18, 19), (18, 19, 20)\}$
$\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}$
$=\frac{18}{^{20}\text{C}_3}$
Required probability $=\frac{18}{^{20}\text{C}_3}$
View full question & answer→Question 1032 Marks
Two cards are drawn without replacement from a pack of 52 cards. Find the probability that.
The first is a king and the second is an ace.
AnswerWe know that, there are 4 king and 4 ace in a pack of 52 cards.
Two cards are drawn without replacment
A = First catd is king
B = Second card an ace
P (The first card is a king second is an ace)
$=\text{P(A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{4}{52}\times\frac{4}{21}$
$=\frac{4}{663}$
Required probability $=\frac{4}{663}$
View full question & answer→Question 1042 Marks
If $\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11}$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{11},$ find
$\text{P}(\text{A}\cap\text{B})$
AnswerGiven,
$\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11},\text{P}(\text{A}\cup\text{B})=\frac{7}{11}$
Since, $\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})$
$=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}$
$\text{P}(\text{A}\cap\text{B})=\frac{4}{11}$
View full question & answer→Question 1052 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
AnswerGiven that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P(A) }\text{P(B)}$
$=\big[1-\text{P(A)}\big]\big[1-\text{P(B)}\big]$
$=(1-0.3)(1-0.6)$
$=0.7\times0.4$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=0.28$
View full question & answer→Question 1062 Marks
A fair die is rolled. Consider events E = $\{1,\ 3,\ 5\},\ \text{F}=\{2,\ 3\}\ \text{and}\ \text{G}=\{2,\ 3,\ 4,\ 5\}.\ \text{Find}:$
$\text{P}(\text{E}|\text{F})\ \text{and}\ \text{P}(\text{F}|\text{E})$
Answer$\text{S}=(1,\ 2,\ 3,\ 4,\ 5,\ 6)\ \Rightarrow\ \ \ \ \ \ \text{n}(\text{S})=6$
$\text{E}=(1,\ 3,\ 5)\ \ \ \ \ \ \ \ \text{F}=(2,\ 3)\ \ \ \ \ \ \ \ (\text{G})=(2,\ 3,\ 4,\ 5)$
$\Rightarrow\ \ \ \ \ \text{n}(\text{E})=3\ \ \ \ \ \ \ \ \text{n}(\text{F})=2\ \ \ \ \ \ \ \text{n}(\text{G})=4$
$\text{P}\left(\text{E}\right)=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{3}{6}\ \ \ \ \ \ \ \ \ \text{P}\left(\text{F}\right)=\frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{2}{6}$
$\text{E}\cap\text{F}=\left(3\right)\ \Rightarrow\ \ \ \ \text{n}\left(\text{E}\cap\text{F}\right)=1$
$\text{P}\left(\text{E}\cap\text{F}\right)=\frac{\text{n}\left(\text{E}\ \cap\ \text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{1}{6}$
$\text{P}\left(\text{E}|\text{F}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{\frac{1}{6}}{\frac{2}{6}}=\frac{1}{2}\ \ \ \ \ \ \ \ \\ \text{and}\ \ \ \text{P}\left(\text{F}|\text{E}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{F}\right)}{\text{P}\left(\text{E}\right)}=\frac{\frac{1}{6}}{\frac{3}{6}}=\frac{1}{3}$
View full question & answer→Question 1072 Marks
A bag contains 7 red, 5 white and 8 black balls. If four balls are drawn one by one with replacement, what is the probability that
none is white?
AnswerLet P be the probability of getting 1 white ball out of 7 red, 5 white and 8 black balls. so
$\text{p}=\frac{5}{20}$
$\text{p}=\frac{1}{4}$
$\text{q}=1-\frac{1}{4}$ [Since p + q = 1]
Let X denote the random variable of number of selecting white ball with replacement out of 4 balls. probability of getting r white balls out of n balls is given by
$\text{P}(\text{x = r})=\text{ }^\text{n}\text{c}_\text{r}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^4\text{c}_{\text{r}}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{4-\text{r}}\dots(1)$
Probabitilty of getting none white ball
$=\text{P}(\text{x}=0)$
$=\text{ }^4\text{c}_0\big(\frac{1}{4}\big)^0\big(\frac{3}4{}\big)^{4-0}$ [Using (1)]
$=\big(\frac{3}{4}\big)^4$
$=\frac{81}{256}$
Probability of getting none white ball $=\frac{81}{256}$
View full question & answer→Question 1082 Marks
In a group of 200 items, if the probability of getting a defective item is 0.2, write the mean of the distribution.
AnswerLet p be the probability of defective item, so
$\text{p}=0.2$
$\text{p}=\frac{1}{5}$
$\text{q}=1-\frac{1}{5}$ [Since p+ q =1]
$\text{q}=\frac{4}{5}$
Given, $\text{n}=200$
$\text{Mean = np}$
$=200\big(\frac{1}{5}\big)$
$=40$
$\text{Mean}=40$
View full question & answer→Question 1092 Marks
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that.
Both the balls are red.
AnswerGiven,
Box contains 10 black and 8 red balls.
Two balls are drawn with replacement.
P (Both the balls are red)
$=\text{P}(\text{R}_1\cap\text{R}_2)$
$=\text{P}(\text{R}_1)\text{ P}(\text{R}_2)$
$=\frac{8}{18}\times\frac{8}{18}$
$=\frac{16}{81}$
Required probability $=\frac{16}{81}$
View full question & answer→Question 1102 Marks
One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
E: ‘the card drawn is a spade’
F: ‘the card drawn is an ace’
AnswerIn a deck of 52 cards, 13 cards are spades and 4 cards are aces.
$\therefore$ P(E) = P(the card drawn is a spade) $=\frac{13}{52}=\frac{1}{4}$
$\therefore$ P(F) = P(the card drawn is an ace)$=\frac{4}{52}=\frac{1}{13}$
In the deck of cards, only 1 card is an ace of spades.
P(EF) = P(the card drawn is spade and an ace) $=\frac{1}{52}$
$\text{P}(\text{E})\times\text{P}(\text{F})=\frac{1}{4}\cdot\frac{1}{13}=\frac{1}{52}=\text{P}(\text{EF})$
$\Rightarrow\text{P}(\text{E})\times\text{P}(\text{F})=\text{P}(\text{EF})$
Therefore, the events E and F are independent.
View full question & answer→Question 1112 Marks
A card is drawn from a well-shulffled deck of 52 cards. The outcome is noted, the card is replaced and the deck reshuffled. Another card is then drawn from the deck.
What is the probability that the first card is an ace and the second card is a red queen?
AnswerP(first ace and second red queen) = P(ace card) × P(red queen)
$=\frac{4}{52}\times\frac{2}{52}$
$=\frac{1}{338}$
View full question & answer→Question 1122 Marks
A discrete random variable X has the probability distribution given below:
| X: |
0.5 |
1 |
1.5 |
2 |
| P(X): |
$k$ |
$k^2$ |
$2k^2$ |
$k$ |
Determine the mean of the distribution. AnswerMean $=\sum\text{p}_\text{i}\text{x}_\text{i}=0.5\times\text{k}+1\times\text{k}^2+1.5\times2\text{k}^2+2\times\text{k}$
$=0.5\times\frac{1}{3}+1\times\Big(\frac{1}{3}\Big)^2+1.5\times2\Big(\frac{1}{3}\Big)^2+2\times\frac{1}{3}$
$=\frac{0.5}{3}+\frac{1}{9}+\frac{3}{9}+\frac{2}{3}$
$=\frac{1.5+1+3+6}{9}$
$=\frac{11.5}{9}$
$=\frac{115}{90}$
$=\frac{23}{18}$
View full question & answer→Question 1132 Marks
Five cards are drawn successively with replacement from a well-shuffled pack of 52 cards. What is the probability that
none is a spade?
AnswerLet P denote the probability of geting one spade out of a deck of 52 cards, So
$\text{p}=\frac{13}{52}$
$\text{p}=\frac{1}{4}$
$\text{q}=1-\frac{1}{4}$ [Since p + q = 1]
$\text{q}=\frac{3}{4}$
let X denote the radom variable of number of spades out of 5 cards. probability of getting r spades out of n cards is given by
$\text{P}(\text{x = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^5\text{c}_\text{r}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{5-\text{r}}\dots(1)$
P(none is a spade) = P(X = 0)
$=\text{ }^5\text{C}_0\big(\frac{1}{4}\big)^0\big(\frac{3}{4}\big)^5$
$=\frac{243}{1024}$
View full question & answer→Question 1142 Marks
If A and B are two events such that $\text{P}(\text{A}\cap\text{B})=0.32$ and P (B) = 0.5, find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$.
AnswerGiven:
$\text{P}(\text{A}\cap\text{B})=0.32$ and $\text{P(B)}=0.5$
We know that,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$=\frac{0.32}{0.5}$
$=\frac{16}{25}$
$=0.64$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.64$
View full question & answer→Question 1152 Marks
A bag contains 4 red and 5 black balls, a second bag contains 3 red and 7 black balls. One ball is drawn at random from each bag, find the probability that the,
Balls are of different colours.
AnswerGiven,
Bag (1) contains 4 red and 6 black balls.
Bag (2) contains 3 red and 7 black balls
One ball is drawn ar random from each bag.
P (Balls are of different colours)
$=\text{P}\big((\text{R}_1\cap\text{B}_2)\cup(\text{B}_1\cap\text{R}_2)\big)$
$=\text{P}(\text{R}_1\cap\text{B}_2)+\text{P}(\text{B}_1\cap\text{R}_2)$
$=\text{P}(\text{R}_1)\text{P}(\text{B}_2)+\text{P}(\text{B}_1)\text{P}(\text{R}_2)$
$=\frac{4}{9}\times\frac{7}{10}+\frac{5}{9}\times\frac{9}{10}$
$=\frac{28}{90}+\frac{15}{90}$
$=\frac{43}{90}$
View full question & answer→Question 1162 Marks
If X follows binomial distribution with parameters n = 5, p and P(X = 2) = 9P(X = 3), then find the value of p.
AnswerWe have,
X follows binomial distribution with parameters $\text{n = 5, p}$ and $\text{P(X}=2)=9\text{P(X}=3).$
So, $\text{P(X = r})=\text{ }^{5}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(5-\text{r})},$ where $\text{r}=0,1,2,3,4,5$ and $\text{q}=1-\text{p}$
As, $\text{P(X}=2)=9\text{P(X}=3)$
$\Rightarrow\text{ }^5\text{c}_2\text{p}^2\text{q}^3=9^5\text{c}_3\text{p}^3\text{q}^2$
$\Rightarrow10\text{p}^2\text{q}^3=9\times10\text{p}^3\text{q}^2$
$\Rightarrow\text{q}=9\text{p}$
$\Rightarrow1-\text{p}=9\text{p}$ [As, q = 1 - p]
$\Rightarrow10\text{p}=1$
$\therefore\text{p}=\frac{1}{10}$
View full question & answer→Question 1172 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
AnswerGiven that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{0.18}{0.3}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.6$
View full question & answer→