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MATHS 2 Marks Questions

Probability · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 117 questions.

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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.
Answer
Event 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.
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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.
$\therefore \text{ 0.6 = 0.4 + p – (0.4)p}$
$\Rightarrow \text{p} = \frac{1}{3}$
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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.
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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.
Answer
Let 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 $\text{E}\cap\text{F}=\{(5,3)(6,2)\}\Rightarrow\text{P}(\text{E}\cap\text{F})=\frac{2}{36}$
$\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)=\frac{\text{P}(\text{E}\cap\text{F})}{\text{P}(\text{F})}=\frac{\frac{2}{36}}{\frac{18}{36}}$
$=\frac{1}{9}$
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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.
Answer
When 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.
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Question 62 Marks
A die is thrown 6 times. If “getting an odd number” is a “success”, what is the probability of:
  1. 5 successes?
  2. Atmost 5 successes?
Answer
Let'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 $\text{P(q)}=\frac{1}{2}$
  1. Probability of getting 5 successes =
$^6\text{C}_5\Big(\frac{1}{2}\Big)^5\Big(\frac{1}{2}\Big)^1=\ ^6\text{C}_5\Big(\frac{1}{2}\Big)^6=\frac{3}{32}.$
  1. Probability of getting atmost 5 successes =
 $\text{P}(\text{p}\leq5)=1-\text{P}(\text{p}=6)$
$=1-\ ^6\text{C}_6\Big(\frac{1}{2}\Big)^6=1-\frac{1}{64}$
$=\frac{63}{64}.$
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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’.
Answer
It is known that the sum of probabilities of a probability distribution of random variables is one.
$\therefore\text{k}+2\text{k}+3\text{k}+0=1$
$\Rightarrow6\text{k}=1$
$\Rightarrow\text{k}=\frac{1}{6}$
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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?
Answer
Given 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 $\Big(\frac{3}{10}\Big)^2\times\frac{7}{10}+\frac{7}{10}\times\Big(\frac{3}{10}\Big)^2$
$= \frac{9}{100}\times\frac{7}{10}+\frac{7}{10}\times\frac{9}{100}$
$= \frac{63}{1000} + \frac{63}{1000}$
$=\frac{63+63}{1000}$
$=\frac{126}{1000}$
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Question 92 Marks
Six coins are tossed simultaneously. Find the probability of getting.
no heads.
Answer
Let 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)
$=\frac{\text{ }^6\text{C}_0}{2^6}$
$=\big(\frac{1}{2}\big)^6$
$=\frac{1}{64}$
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Question 102 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
$\text{X}$ $\text{P(X)}$
$0$ $\frac{7}{6}\times\frac{6}{8}=\frac{21}{36}$
$1$ $\frac{7}{9}\times\frac{2}{8}\times2=\frac{14}{36}$
$2$ $\frac{2}{9}\times\frac{1}{8}=\frac{1}{36}$
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Question 112 Marks
Find the probability distribution of the number of sixes in three tosses of a die.
Answer
$\text{X}$ $\text{P(X)}$
$0$ $\text{}^3\text{c}_0\big(\frac{1}{6}\big)^0\big(\frac{5}{6}\big)^{3-0}=\big(\frac{5}{6}\big)^3=\frac{125}{216}$
$1$ $\text{ }^3\text{c}_1\big(\frac{1}{6}\big)^1\big(\frac{5}{6}\big)^{3-1}=3\big(\frac{1}{6}\big)\big(\frac{1}{6}\big)^2=\frac{25}{72}$
$2$ $\text{ }^3\text{c}_2\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^{3-2}=3\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)=\frac{5}{72}$
$3$ $\text{ }^3\text{c}_3\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)^{3-3}=\big(\frac{1}{6}\big)^3=\frac{1}{216}$
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Question 122 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.
Answer
Let $A$ 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.
$\text{P}(\text{E}_1)=\frac{1}{2},\ \text{P}(\text{E}_2)=\frac{1}{2},$
$\text{P}(\text{A}|\text{E}_1)=\text{P}(\text{drawing a red ball from bag I})=\frac{4}{8}=\frac{1}{2}$
$\text{P}(\text{A}|\text{E}_2)=\text{P}(\text{drawing a red ball from bag II})=\frac{2}{8}=\frac{1}{4}$
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Question 132 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 $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{1}{4}+\frac{2}{8}+\frac{15}{8}=\frac{2+2+15}{8}=\frac{19}{8}$
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Question 142 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).
Answer
A and B are two independent events
$\therefore\ \text{P}(\text{A}\cap\overline{\text{B}})=\text{P(A)}+\text{P}(\overline{\text{B}})-\text{P}(\text{A}\cap\overline{\text{B}})$
$\Rightarrow\ 0.8=0.3+[1-\text{P(B)}]-\text{P(A)}\text{P}(\overline{\text{B}})$
$\Rightarrow 0.5=1-\text{P(B)}-0.3[1-\text{P(B)}]$
$\Rightarrow 0.5=1-\text{P(B)}=0.3+0.3\text{P(B)}$
$\Rightarrow\ 0.5= 0.7-\text{P(B)}[1-0.3]$
$\Rightarrow\ 0.7\text{P(B)} = 0.2$
$\Rightarrow\ \text{P(B)}=\frac{0.2}{0.7}=\frac{2}{7}$
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Question 152 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?
Answer
It 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.
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Question 162 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}$
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Question 172 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.
Answer
We 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}$
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Question 182 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}$
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Question 192 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}$
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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.
Answer
Total 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}$
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Question 212 Marks
A die is tossed thrice. Find the probability of getting an odd number at least once.
Answer
Probability 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}$
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Question 222 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.
Answer
Bag 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}$
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Question 232 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.
Answer
Bag 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}$
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Question 242 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)$
Answer
Given,
$\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}$
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Question 252 Marks
6 boys and 6 girls sit in a row at random. Find the probability that all the girls sit together.
Answer
Here, 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}$
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Question 262 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%?
Answer
Let 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.
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Question 272 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.
Answer
There 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}$
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Question 282 Marks
Write the probability that a number selected at random from the set of first 100 natural numbers is a cube.
Answer
Number 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}$
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Question 292 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.
Answer
Given,
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}$
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Question 302 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.
Answer
Let 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}$
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Question 312 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.
Answer
Here = 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$
$=\text{p}^4\big[\ ^6\text{C}_2\ \text{q}^2+\ ^6\text{C}_1\ \text{p}\ \text{q}+\ ^6\text{C}_0\ \text{p}^2\big]$
$=\bigg(\frac{2}{3}\bigg)^4\bigg\{\frac{6\times5}{1\times2}\bigg(\frac{1}{3}\bigg)^2+\frac{6}{1}.\frac{2}{3}.\frac{1}{3}+1\times\bigg(\frac{2}{3}\bigg)^2\bigg\}$
$=\frac{16}{81}\bigg\{\frac{15}{9}+\frac{12}{9}+\frac{4}{9}\bigg\}=\frac{16}{81}\times\frac{31}{9}=\frac{496}{729}.$
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Question 322 Marks
Suppose X has a binomial distribution with n = 6 and p = $\frac{1}{2}.$ Show that X = 3 is the most likely outcome.
Answer
we 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
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Question 332 Marks
Two cards are drawn without replacement from a pack of 52 cards. Find the probability that.
Both are kings.
Answer
In 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 $=\frac{1}{221}$
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Question 342 Marks
If A and B are two events write the expression for the probability of occurrence of exactly one of two events.
Answer
P(exaxtly one of 2 events) $=\text{P}(\text{A}\cup\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\text{P(A)}+\text{P(B)}-2\text{P}(\text{A}\cap\text{B})$
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Question 352 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.
Answer
Die 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
$\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}$
$=\frac{4\times4\times4\times4}{6\times6\times6\times6}$
$=\frac{16}{81}$
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Question 362 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}$
  1. Determine the value of $k.$
  2. Find $P(X < 2), P(X ≤ 2), P(X ≥ 2).$
Answer
Probability distribution:
$x_i$ $0$ $1$ $2$
$P(x_i)$ $k$ $2k$ $3k$
  1. $P(X = 0) + P(X = 1) + P(X = 2) = 1$
$\Rightarrow k + 2k + 3k = 1 $
$\Rightarrow 6k = 1$
$\Rightarrow\ \text{k}=\frac{1}{6}$
  1. $P(X < 2) = P(X = 0) + P(X = 1)$
$=\text{k}+2\text{k}=3\text{k}=3\times\frac{1}{6}=\frac{1}{2}$
$P(X \geq 2) = P(X = 2) = 3k $
$=3\times\frac{1}{6}=\frac{1}{2}$
 
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Question 372 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 $\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big).$
Answer
According 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, $\text{P}(\overline{\text{B}}\cup(\text{A}\cup\text{B}))=\phi$
$\therefore\ \text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big)=0$
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Question 382 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
Answer
P(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.
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Question 392 Marks
When three dice are thrown, write the probability of getting 4 or 5 on each of the dice simultaneously.
Answer
There 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 $=\frac{1}{27}$
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Question 402 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\text{A}\cup\text{B})$
Answer
Given that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\text{A}\cup\text{B})=\text{P(A)} + \text{P(B)} - \text{P}(\text{A}\cap\text{B})$
$=0.3+0.6-0.18$
$\text{P}(\text{A}\cup\text{B})=0.72$
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Question 412 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.
Answer
Here,
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
$\Rightarrow\text{a}=\frac{1}{81}$
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Question 422 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’
Answer
In 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.
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Question 432 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\text{A}\cap\text{B})$
Answer
Given 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]
$=0.3\times0.6$
$\text{P}(\text{A}\cap\text{B})=0.18$
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Question 442 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
Answer
P(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.
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Question 452 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.
Answer
There 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 $=\frac{25}{204}$
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Question 462 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:
$(1 - p_1)p_2.$
Answer
As, $(1 - p_1)p_2$
$=[1-\text{P(A)}]\times\text{P(B)}=\text{P}(\overline{\text{A}})\times\text{P(B)}$
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$).$
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Question 472 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.
Answer
Given,
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 $=\frac{47}{90}$
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Question 482 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\text{A}\cap\overline{\text{B}})$
Answer
Given that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\text{A}\cap\overline{\text{B}})=\text{P(A)}-\text{P}(\text{A}\cap\text{B})$
$=0.3 - 0.18$
$\text{P}(\text{A}\cap\overline{\text{B}})=0.12$
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Question 492 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.
Answer
An 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 $=\frac{49}{121}$
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Question 502 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
$\text{F}=(\text {MFS, SFM})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \text{n}(\text{F})=2$
$\text{P}\left (\text{F}\right)=\frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{2}{6}=\frac{1}{3}$
$\therefore\ \ \ \ \text{E}\cap\text{F}=\left(\text{MFS, SFM}\right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \text{n}\left(\text {E}\cap\text{F}\right)=2$
$\therefore\ \ \ \ \text{P}\left(\text{E}\cap\text{F}\right)=\frac{\text{n}\left(\text{E}\ \cap\ \text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{2} {6}=\frac{1}{3}$
$\text{And}\ \ \ \ \text{P}\left(\text{E}|\text{F}\right)=\frac{\text{P}\left(\text{E}\ \cap\ \text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{\frac {1}{3}}{\frac{1}{3}}=1$
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2 Marks Questions - MATHS STD 12 Science Questions - Vidyadip