MCQ 11 Mark
If A and B are two independent events with $\text{P(A)}=\frac{1}{3}$ and $\text{P(B)}=\frac{1}{4},$ then P(B'|A) is equal to:
- A
$\frac{1}{4}$
- B
$\frac{1}{3}$
- ✓
$\frac{3}{4}$
- D
$1$
AnswerCorrect option: C. $\frac{3}{4}$
$\text{P(A)}=\frac{1}{3},\text{ P(B)}=\frac{1}{4},\text{ P}\Big(\frac{\text{B}'}{\text{A}}\Big)=?$
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P(B)}$
$\text{P}\Big(\frac{\text{B}'}{\text{A}}\Big)=\text{P}\frac{(\text{A}\cap\text{B}')}{\text{P(A)}}$
$\text{P}(\text{B}')=1-\frac{1}{4}=\frac{3}{4}$
$\therefore\text{P}\Big(\frac{\text{B}'}{\text{A}}\Big)=\frac{\text{P(A)}\times\text{P(B})'}{\text{P(A)}}$
$=\text{P(B}')=\frac{3}{4}$
View full question & answer→MCQ 21 Mark
Choose the correct answer from the given four options.
In a college, 30% students fail in physics, 25% fail in mathematics and 10% fail in both. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is:
- A
$\frac{1}{10}$
- ✓
$\frac{2}{5}$
- C
$\frac{9}{20}$
- D
$\frac{1}{3}$
AnswerCorrect option: B. $\frac{2}{5}$
Here, $\text{P}_{(\text{Ph})}=\frac{30}{100}=\frac{3}{10}$
$\text{P}_{(\text{M})}=\frac{25}{100}=\frac{1}{4}$
And $\text{P}_{(\text{M}\cap\text{Ph})}=\frac{10}{100}=\frac{1}{10}$
$\therefore\text{P}\Big(\frac{\text{Ph}}{\text{M}}\Big)=\frac{\text{P}(\text{Ph}\cap\text{M})}{\text{P}(\text{M})}$
$=\frac{\frac{1}{10}}{\frac{1}{4}}=\frac{2}{5}$
View full question & answer→MCQ 31 Mark
If X is a binomial variate with parameters n and p, where 0 < p < 1 such that $\frac{\text{P(X = r)}}{\text{P(X = n - r})}$ is independent of n and r, then p equals:
- ✓
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{1}{4}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\frac{1}{2}$
Consider,
$\text{P(X = r})=\text{kP(X = n}-\text{r})$
Using $\text{ }^{\text{n}}\text{C}_{\text{r}}=\text{ }^{\text{n}}\text{C}_{\text{n}-\text{r}},\text{q}=1-\text{p}$
$\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}=\text{kp}^{\text{n}-\text{r}}\text{q}^{\text{r}}$
$\text{p}^{\text{r}}(1-\text{p})^{\text{n}-\text{r}}=\text{kp}^{\text{n}-\text{r}}(1-\text{p})^{\text{r}}$
$\text{P}^{2\text{r}-\text{n}}=\text{k}(1-\text{p})^{2\text{r}-\text{n}}$
$\big(\frac{\text{p}}{\text{q}}\big)^{2\text{r}-\text{n}}=\text{k}$
when $\text{p = q}$ then $\text{k}=1$
$\Rightarrow\text{p = q}=\frac{1}{2}$
View full question & answer→MCQ 41 Mark
Choose the correct answer in each of the following:
Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is
- A
$\frac{37}{221}$
- B
$\frac{5}{13}$
- ✓
$\frac{1}{13}$
- D
$\frac{2}{13}$
AnswerCorrect option: C. $\frac{1}{13}$
$\text{n}(\text{S})=52,\ \text{n}(\text{A})=4$
$\text{P}(\text{X}=0)=\frac{^{48}\text{C}_2}{^{52}\text{C}_2}=\frac{48\times47}{52\times51}=\frac{188}{221}$
$\text{P}(\text{X}=1)=\frac{^{48}\text{C}_2\times^4\text{C}_1}{^{52}\text{C}_2}=\frac{2\times48\times4}{52\times51}=\frac{32}{221}$
$\text{P}(\text{X}=2)=\frac{^4\text{C}_2}{^{52}\text{C}_2}=\frac{4\times3}{52\times51}=\frac{1}{221}$
| $\text{x}_i$ |
$\text{p}_i$ |
$\text{p}_i\text{x}_i$ |
$0$
$1$
$2$ |
$\frac{188}{221}$
$\frac{32}{221}$
$\frac{1}{221}$ |
$0$
$\frac{32}{221}$
$\frac{2}{221}$ |
| |
|
$\sum\text{p}_i\text{x}_i=\frac{34}{221}=\frac{2}{13}$ |
$\text{Now}\ \text{E}(\text{X})=\frac{2}{13}$
Therefore, option (D) is correct. View full question & answer→MCQ 51 Mark
If P(A) + P(B) = 1; then which of the following option explains the event A and B correctly?
- ✓
Event A and B are mutually exclusive, exhaustive and complementary events.
- B
Event A and B are mutually exclusive and exhaustive events.
- C
Event A and B are mutually exclusive and complementary events.
- D
Event A and B are exhaustive and complementary events.
AnswerCorrect option: A. Event A and B are mutually exclusive, exhaustive and complementary events.
Since P(A) + P(B) = 1
$\therefore\text{A}\cap\text{B}=0.$
Thus, event A and B are mutually exclusive, exhaustive and complementary events.
View full question & answer→MCQ 61 Mark
In a box containing 100 bulbs, 10 are defective. What is the probability that out of a sample of 5 bulbs, none is defective?
AnswerCorrect option: A. $\big(\frac{9}{10}\big)^5$
Let X denote the number of defective bulbs.
Hence, the binomial distribution is given by
$\text{n}=5,\text{p}=\frac{10}{100}=\frac{1}{10}$
$\& \text{ q}=\frac{90}{100}=\frac{9}{10}$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{10}\big)^{\text{r}}\big(\frac{9}{10}\big)^{5-\text{r}}$
$\therefore\text{P(X}=0)=\big(\frac{9}{10}\big)^5$
View full question & answer→MCQ 71 Mark
Choose the correct answer from the given four options.
| $\text{X}$ |
$1$ |
$2$ |
$3$ |
$4$ |
| $\text{P}(\text{X})$ |
$\frac{1}{10}$ |
$\frac{1}{5}$ |
$\frac{3}{10}$ |
$\frac{2}{5}$ |
For the following probability distribution $E(X^2)$ is equal to: Answer$\text{E}(\text{X}^2)=\sum\text{X}^2\text{P}(\text{X})$
$=1\cdot\frac{1}{10}+4\cdot\frac{1}{5}+9\cdot\frac{3}{10}+16\cdot\frac{2}{5}$
$=\frac{1}{10}+\frac{4}{5}+\frac{27}{10}+\frac{32}{5}$
$=\frac{1+8+27+64}{10}=10$
View full question & answer→MCQ 81 Mark
Choose the correct answer from the given four options.
If A and B are two independent events with $\text{P}(\text{A})=\frac{3}{5}$ and $\text{P}(\text{A})=\frac{4}{9},$ then $\text{P}(\text{A'}\cap\text{B'})$ equals:
- A
$\frac{4}{15}$
- B
$\frac{8}{45}$
- C
$\frac{1}{3}$
- ✓
$\frac{2}{9}$
AnswerCorrect option: D. $\frac{2}{9}$
Since A and B are independent events, A' And B' are aslo independent.
$\therefore\text{P}(\text{A}'\cap\text{B}')=\text{P}(\text{A})\cdot\text{P}(\text{B})$
$=\big[1-\text{P}(\text{A})\big]\big[1-\text{P}(\text{B})\big]$
$=\Big(1-\frac{3}{5}\Big)\Big(1-\frac{4}{9}\Big)$
$=\frac{2}{5}\cdot\frac{5}{9}=\frac{2}{9}$
View full question & answer→MCQ 91 Mark
Choose the correct answer from the given four options.
|
X
|
-4
|
-3
|
-2
|
-1
|
0
|
|
P(X)
|
0.1
|
0.2
|
0.3
|
0.2
|
0.2
|
For the following probability distribution E(X) is equal to:
Answer$\text{E}(\text{X})=\sum\text{P}(\text{X})$
$= -4\times(0.1)+(-3 \times0.2)+(-2\times0.3)+(-1\times0.2)+(0\times0.2)$
$=-0.4-0.6-0.6-0.2=-1.8$
View full question & answer→MCQ 101 Mark
A five-digit number is written down at raddom. The probability that the number is divisible by 5, and no two consecutive digits are identical, is:
AnswerCorrect option: D. $\text{None of these}$
If last digit is either O or 5 then the number is divisible by 5.
Case : 1
Last digit is 0.
First three places can be selected by 9 × 9 × 9 = 729 ways.
If c = 0 then three places can be selected by 9 × 8 × 1 = 72
If C ≠ 0 then 729 - 72 = 657
Fourth place has 8 choices = 657 × 8 = 5256
Total = 72 + 5256 = 5904
Case : 2
If C = 5
First place other than 5
then first three places can be filled in 8 × 8 × 1 = 64
If first place is 5 then first three places can be filled in 1× 9 × 1 = 9 ways.
If third place is other than 5 then 729 - 64 - 9 = 656 ways.
For fourth place has 8 choices.
As per required condition = (64 + 9) × 9 + 656 × 8 = 5905
required probability $=\frac{5904+5905}{9\times10\times10\times10\times10}=\frac{11809}{90000}$
NOTE: Answer not matching with back answer.
View full question & answer→MCQ 111 Mark
If A and B are two events such that $\text{P(A)}=\frac{3}{8},\text{P(B)}=\frac{5}{4}.$ and $\text{P}(\text{A}|\text{B})\times\text{P}(\overline{\text{A}}\cap\text{B})$ is equals to.
- A
$\frac{2}{5}$
- B
$\frac{3}{8}$
- C
$\frac{3}{20}$
- ✓
$\frac{6}{25}$
AnswerCorrect option: D. $\frac{6}{25}$
$\text{P(A)}=\frac{3}{8},\text{P(B)}=\frac{5}{8},\text{P}\Big({\text{A}}\cup{\text{B}}\Big)=\frac{3}{4}$
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \frac{3}{4}=\frac{3}{8}+\frac{5}{8}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{4}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}\times\frac{\text{P}(\overline{\text{A}}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}\times\frac{\text{P(B)}-\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\frac{1}{4}}{\frac{5}{8}}\times\frac{\big(\frac{5}{8}-\frac{1}{4}\big)}{\frac{5}{8}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{6}{25}$
View full question & answer→MCQ 121 Mark
If the mean and variance of a binomial distribution are 4 and 3, respectively, the probability of getting exactly six successes in this distribution is:
- A
$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{10}\big(\frac{3}{4}\big)^6$
- ✓
$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{6}\big(\frac{3}{4}\big)^{10}$
- C
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)\big(\frac{3}{4}\big)^6$
- D
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)^6\big(\frac{3}{4}\big)^6$
AnswerCorrect option: B. $\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{6}\big(\frac{3}{4}\big)^{10}$
$\text{np}=4,\text{npq}=3$
$\Rightarrow\text{q}=\frac{3}4{},\text{p}=\frac{1}{4},\text{n}=16$
$\text{P(X}=6)=\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^6\big(\frac{3}{4}\big)^{10}$
View full question & answer→MCQ 131 Mark
Choose the correct answer from the given four options.
If the events A and B are independent, then $\text{P}(\text{A}\cap\text{B})$ is equal to:
- A
$\text{P}(\text{A})+\text{P}(\text{B})$
- B
$\text{P}(\text{A})-\text{P}(\text{B})$
- ✓
$\text{P}(\text{A})\cdot\text{P}(\text{B})$
- D
$\frac{\text{P}(\text{A})}{\text{P}(\text{B})}$
AnswerCorrect option: C. $\text{P}(\text{A})\cdot\text{P}(\text{B})$
If A and B are independent, then $\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})\cdot\text{P}(\text{B})$
View full question & answer→MCQ 141 Mark
10 persons are seated around a round table. What is the probability that 4 particular persons are always seated together?
- A
$\frac{1}{20}$
- B
$\frac{4}{10}$
- ✓
$\frac{1}{21}$
- D
$\frac{3}{20}$
AnswerCorrect option: C. $\frac{1}{21}$
10 persons can sit around a table in 9! ways.
Consider the particular four persons as one unit.
Now, the entities are 6 + 1 = 7
These 7 entities can be arranged in 6! ways.
In the entities itself they can be arranged in 4! ways.
The required number of arrangements = 6!4!
Probability $= \text{nm} = \frac{!6!4}{9!} = \frac1{21}$
View full question & answer→MCQ 151 Mark
If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is:
- A
$\frac{2}{3}$
- B
$\frac{4}{5}$
- C
$\frac{7}{8}$
- ✓
$\frac{15}{16}$
AnswerCorrect option: D. $\frac{15}{16}$
$\text{E(X)}=2, \text{V(X})=1$
$\text{np}=2,\text{npq}=1$
$\Rightarrow\text{q}=\frac{1}2{}=\text{p}$
$\Rightarrow\text{n}=4$
$\text{P(X}\geq1)=1-\text{P(X}=0)$
$\text{P(X}\geq1)=1-\big(\frac{1}{2}\big)^4$
$\text{P(X}\geq1)=1-\frac{1}{16}=\frac{15}{16}$
View full question & answer→MCQ 161 Mark
A bag contains 5 brown and 4 white socks. A man pulls out two socks. The probability that these are of the sane colour is.
- A
$\frac{5}{108}$
- B
$\frac{18}{108}$
- ✓
$\frac{30}{108}$
- D
$\frac{48}{108}$
AnswerCorrect option: C. $\frac{30}{108}$
Total number of balls = 5brown + 4white = 9
Required probability $=\frac{5}{9}\times\frac{4}{8}+\frac{4}{9}\times\frac{3}{8}=\frac{4}{9}$
$\Rightarrow\ \frac{4\times12}{9\times12}=\frac{48}{108}$
View full question & answer→MCQ 171 Mark
The least number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8, is:
AnswerA fair coin is tossed $\Rightarrow\text{p = q}=\frac{1}{2}$
$\text{P(X}\geq1)\geq0.8$
$\Rightarrow1-\text{P}(0)\geq0.8$
$\Rightarrow\text{P}(0)=0.2$
$\Rightarrow\big(\frac{1}{2}\big)^{\text{n}}=0.2$
$\Rightarrow2^{-\text{n}}=0.2$
$\Rightarrow2^{\text{n}}\geq5$
$\Rightarrow\text{n}\geq3$
View full question & answer→MCQ 181 Mark
An urn contains 9 balls two of which are red, three blue and four black. Three balls are drawn at random. The probability that they are of the same colour is,
- ✓
$\frac{5}{84}$
- B
$\frac{3}{9}$
- C
$\frac{3}{7}$
- D
$\frac{7}{17}$
AnswerCorrect option: A. $\frac{5}{84}$
Given:
Red balls = 2
Blue balls = 3
Black balls = 4
P(All three balls are of same colour) = P(all three are blue) + P(all three are black)
$=\frac{3}{9}\times\frac{2}{8}\times\frac{1}{7}+\frac{4}{9}\times\frac{3}{8}\times\frac{2}{7}$
$=\frac{1}{84}+\frac{4}{84}$
$=\frac{5}{84}$
View full question & answer→MCQ 191 Mark
A four - digit number is formed by using the digits 1, 2, 4, 8 and 9 without repitition. If one number is selected from those numbers, then what is the probability that it will be an odd number?
- A
$\frac{1}{5}$
- ✓
$\frac{2}{5}$
- C
$\frac{3}{5}$
- D
$\frac{4}{5}$
AnswerCorrect option: B. $\frac{2}{5}$
Total number of outcomes = 5 × 4 × 3 × 2 = 120
he number of favourable cases = 2 (4 × 3 × 2) - = 48 (i.e., odd numbers)
herefore,Required probability $=\frac{48}{120}=\frac{2}{5}.$
View full question & answer→MCQ 201 Mark
Choose the correct answer in each of the following:
The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
Answer
| $\text{x}_i$ |
$\text{p}_i$ |
$\text{p}_i\text{x}_i$ |
$1$
$2$
$5$ |
$\frac{3}{6}$
$\frac{2}{6}$
$\frac{1}{6}$ |
$\frac{3}{6}$
$\frac{4}{6}$
$\frac{5}{6}$ |
| |
|
$\sum\text{p}_i\text{x}_i=\frac{12}{6}=2$ |
Therefore, option (B) is correct. View full question & answer→MCQ 211 Mark
Choose the correct answer from the given four options.
Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If $\frac{\text{P}(\text{x}=\text{r})}{\text{P}(\text{x}=\text{n}–\text{r})}$ is independent of n and r, then p equals:
- ✓
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{1}{5}$
- D
$\frac{1}{7}$
AnswerCorrect option: A. $\frac{1}{2}$
$\text{P}(\text{x}=\text{r})={^\text{n}}\text{C}_\text{r}(\text{p})^\text{r}(\text{q})^{\text{n}-\text{r}}$
$=\frac{\text{n}!}{(\text{n}-\text{r})!\text{r}!}(\text{p})^\text{r}(1-\text{p})^{\text{n}-\text{r}}[\therefore\text{q}=1-\text{p}]$
Now, $\frac{\text{P}(\text{x}=\text{r})}{\text{P}(\text{x}=\text{n}-\text{r})}=\frac{{^\text{n}\text{C}_\text{r}\text{p}^\text{r}(1-\text{p})^{\text{n}-\text{r}}}}{{{^\text{n}}\text{C}}_{\text{n}-\text{r}}\text{p}^{\text{n}-\text{r}}(1-\text{p})^{\text{r}}}$
$=\frac{\text{P}^\text{r}(1-\text{P})^{\text{n}-\text{r}}}{\text{p}^{\text{n}-\text{r}}(1-\text{p})^\text{r}}$ $\big[\text{as}{^\text{n}}\text{C}_\text{r}={^\text{n}}\text{C}_{\text{n}-\text{r}}\big]$
$=\Big(\frac{1-\text{p}}{\text{p}}\Big)^{\text{n}-2\text{r}}$
Above expression is independent of n and r, if
$\frac{1-\text{p}}{\text{p}}=1\Rightarrow\text{p}=\frac{1}{2}$
View full question & answer→MCQ 221 Mark
A die is thrown and a card is selected ar random from a deck $\text{pf 52}$ playing cards. The probability of getting an even number of the die and a spade card is
- A
$\frac{1}{2}$
- B
$\frac{1}{4}$
- ✓
$\frac{1}{8}$
- D
$\frac{3}{4}$
AnswerCorrect option: C. $\frac{1}{8}$
A Sample space when a die is thrown,
$S_1 = \{1, 2, 3, 4, 5, 6\}$
$ \Rightarrow n(S_1) = 6$
Let $A$ be the event that getting even number.
$A = \{2, 4, 6\}$
$\Rightarrow n(A) = 3$
$\Rightarrow\ \text{P(A)}=\frac{3}{6}=\frac{1}{2}$
A card is selected from a deck of $52$ cards.
$\text{n}(\text{S}_2)= {^{52}}\text{C}_2=52$
Let $B$ be the event that getting spade card.
$\text{n(B)}= {^{13}}\text{C}_2=13$
$\Rightarrow\ \text{P(B)}=\frac{13}{52}=\frac{1}{4}$
Required probability $= P(A) \times P(B)$
$=\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}$
View full question & answer→MCQ 231 Mark
Five persone entered the lift cabin on the ground floor of an $8$ floor house. Suppose that each of them independently and with equal probability can leave the cabin at any flor beginning with the first, then the probability of all $5$ persons leaving at different floors is,
- ✓
$\frac{^{7}\text{P}_5}{7_5}$
- B
$\frac{7_5}{^{7}\text{P}_5}$
- C
$\frac{6}{^{6}\text{P}_5}$
- D
$\frac{^{5}\text{P}_5}{5}$
AnswerCorrect option: A. $\frac{^{7}\text{P}_5}{7_5}$
Five persons can leave different floors
By $^7P_5$ ways.
Possible ways of leavinf the lift $= 7^5$
Required probability $=\frac{^{7}\text{P}_5}{7^5}$
View full question & answer→MCQ 241 Mark
A fair coin is tossed 100 times. The probability of getting tails an odd nimber of times is:
- ✓
$\frac{1}{2}$
- B
$\frac{1}{8}$
- C
$\frac{3}{8}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\frac{1}{2}$
Here, $\text{n}=100$
Let X denote the number of times a tail is obtained.
Here, $\text{p = q}=\frac{1}{2}$
$\text{P(X = odd})=\text{P(X}=1,3,5,\dots99)$
$=\big(\text{ }^{100}\text{C}_1+\text{ }^{100}\text{C}_3+\dots+\text{ }^{100}{\text{C}}_{99}\big)\big(\frac{1}{2}\big)^{100}$
= Sum of odd coefficients in binomial expansion in $(1+\text{x})^{100}\big(\frac{1}{2}\big)^{100}$
$=\frac{2^{(100-1)}}{2^{100}}$
$=\frac{1}{2}$
View full question & answer→MCQ 251 Mark
A coin is tossed 4 times. The probability that at least one head turns up is:
- A
$\frac{1}{16}$
- B
$\frac{2}{16}$
- C
$\frac{14}{16}$
- ✓
$\frac{15}{16}$
AnswerCorrect option: D. $\frac{15}{16}$
$\text{n}=4,\text{p = q}=\frac{1}{2}$
$\text{P(X}\geq1)=1-\text{P(X}=0)$
$\text{P(X}\geq1)=1-\big(\frac{1}{2}\big)^4$
$\text{P(X}\geq1)=\frac{15}{16}$
View full question & answer→MCQ 261 Mark
$\int\limits^1_0\sqrt{\text{x}(1-\text{x})}\text{ dx}$ equals:
- A
$\frac{\pi}{2}$
- B
$\frac{\pi}{4}$
- C
$\frac{\pi}{6}$
- ✓
$\frac{\pi}{8}$
AnswerCorrect option: D. $\frac{\pi}{8}$
$\text{I}=\int\limits^1_0\sqrt{\text{x}(1-\text{x})}\text{ dx} $
$\text{I}= \int\limits^1_0\sqrt{\text{x}-\text{x}^2}\text{dx}$
$\text{I}=\int\limits^1_0\sqrt{\frac{1}{4}+\text{x}-\text{x}^2+\frac{1}{4}}\text{dx}$
$\text{I}=\int\limits^1_0\sqrt{\frac{1}{4}-\Big(\text{x}^2-\text{x}+\frac{1}{4}\Big)}\text{dx}$
$\text{I}=\int\limits^1_0\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2}\text{dx}$
$\text{I}=\Bigg[\frac{\text{x}-\frac{1}{2}}{2}\sqrt{\text{x}(1-\text{x})}+\frac{1}{2}\times\frac{1}{4}\sin^{-1}(2\text{x}-1)\Bigg]^1_0$
$\text{I}=0+\frac{1}{8}\big(\sin^{-1}(1)-\sin^{-1}(-1)\big)$
$\text{I}= \frac{1}{8}\Big(\frac{\pi}{2}-\Big(\frac{\pi}{2}\Big)\Big)$
$\text{I}= \frac{\pi}{8}$
View full question & answer→MCQ 271 Mark
Choose the correct answer in each of the following:
If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1, then
AnswerCorrect option: A. $\text{A}\subset\text{B}$
$\text{A}\subset\text{B}$
$\text{P}(\text{B|A})=1$
$\Rightarrow\ \frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}=1\ \ \text{P}(\text{B}\cap\text{A})=\text{P}(\text{A})$
$\therefore$ (A) is correct answer.
View full question & answer→MCQ 281 Mark
Three persons, A, B and C fine a target in turn starting with A. Their probability of hitting the target are 0.4, 0.2 and 0.2, respectively. The probability of two hits is
AnswerLet:
A be the event of hitting the target by the person A,
B be the event of hitting the target by the person B and
C be the event of hitting the target by the person C
We have,
P(A) = 0.4, P(B) = 0.3 and P(C) = 0.2
Also,
$\text{P}(\overline{\text{A}})=1-\text{P(A)}=1-0.4=-0.6,$
$\text{P}(\overline{\text{B}})=1-0.3=0.7$ and
$\text{P}(\overline{\text{C}})=1-0.2=0.8$
Now,
$\text{P(Two hits)}=\text{P}(\text{AB}\overline{\text{C}})+\text{P}(\text{A}\overline{\text{B}}\text{C})+\text{P}(\overline{\text{A}}\text{BC})$
$=\text{P(A)}\times\text{P(B)}\times\text{P}(\overline{\text{C}})+\text{P(A)}\times(\overline{\text{B}})\times\text{P(C)}\\+\text{P}(\overline{\text{A}})\times\text{P(B)}\times\text{P(C)}$
$=0.4\times0.3\times0.8+0.4\times0.7\times0.2+0.6\times0.3\times0.2$
$=0.096+0.056+0.036$
$=0.188$
Hence, the correct alternative is option (d).
View full question & answer→MCQ 291 Mark
Difference between sample space and subset of sample space is considered as:
- A
Numerical complementary events.
- B
- ✓
- D
AnswerThe set of all the possible outcomes is called the sample space of the experiment and is usually denoted by S.
Any subset E of the sample space S
Difference between sample space and subset of sample space is considered as complementary events.
View full question & answer→MCQ 301 Mark
If two events are independent, then.
- A
They must be mutually exclusive.
- B
The sum of their probabilities must be equal to 1.
- C
(a) and (b) both are correct.
- ✓
None of the above is correctIf two. events are independent, then.
AnswerCorrect option: D. None of the above is correctIf two. events are independent, then.
Let A and B are two independent events, Then,
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P(B)}$
As, $\text{P}(\text{A}\cap\text{B})\neq0\text{ or }\text{P(A)}+\text{P(B)}\neq1$
So, both are neither mutually exclisive nor their sum of probability is 1.
Hence, the correct alternative is option (d).
View full question & answer→MCQ 311 Mark
Choose the correct answer from the given four options.
A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, the probability that both are dead is:
- A
$\frac{33}{56}$
- B
$\frac{9}{64}$
- C
$\frac{1}{14}$
- ✓
$\frac{3}{28}$
AnswerCorrect option: D. $\frac{3}{28}$
Required probability $=\text{P}_{\text{D}}\cdot\text{P}_{\text{D}}$
$=\frac{3}{8}\cdot\frac{2}{7}=\frac{3}{28}$
View full question & answer→MCQ 321 Mark
$A$ and $B$ are two students. Their chances of solving a problem correctly are $\frac{1}{3}$ and $\frac{1}{4}$ respectively. If the probability of their making common error is $\frac{1}{20}$ and they obtain the same answer, then the probability of their answer to be correct is.
- ✓
$\frac{10}{13}$
- B
$\frac{13}{120}$
- C
$\frac{1}{40}$
- D
$\frac{1}{12}$
AnswerCorrect option: A. $\frac{10}{13}$
Let $E_1$ be the event that Both $A$ and $B$ solve the problem.
$A$ and $B$ are independent,
$\Rightarrow\ \text{P}(\text{E}_1)=\text{P(A)}\times\text{P(B)}$
$\Rightarrow\ \text{P}(\text{E}_1)=\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$
Let $E_2$ both $A$ and $B$ got wrong solution.
$\text{P}(\text{E}_2)=\Big(1-\frac{1}{3}\Big)\times\Big(1-\frac{1}{4}\Big)=\frac{1}{2}$
Let $E$ be the event getting same answer.
$\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=1,\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{1}{20}$
$\Rightarrow\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=\frac{\frac{1}{12}\times1}{\frac{1}{12}\times1+\frac{1}{2}\times\frac{1}{20}}=\frac{10}{13}$
View full question & answer→MCQ 331 Mark
If A and B are two events such that $\text{P(A)}\neq0$ and $\text{P(B)}\neq1,$ then $\text{P}(\overline{\text{A}}|\overline{\text{B}})=$
- A
$1-\text{P}(\text{A}|\text{B})$
- B
$1-\text{P}(\overline{\text{A}}|\text{B})$
- ✓
$\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$
- D
$=\frac{\text{P}(\overline{\text{A}})}{\text{P}(\overline{\text{B}})}$
AnswerCorrect option: C. $\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$
We have,
$\text{P(A)}\neq0$ and $\text{P(B)}\neq1$
Now,
$\text{P}(\overline{\text{A}}|\overline{\text{B}})=\frac{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}{\text{P}(\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{A}\cap\text{B}})}{\text{P}(\overline{\text{B}})}$
$=\frac{1-\text{P}(\text{A}\cup\text{B})}{\text{P}(\overline{\text{B}})}$
Hence, the correct alternative is option (C).
View full question & answer→MCQ 341 Mark
Choose the correct answer from the given four options.
The probability of guessing correctly at least 8 out of 10 answers on a true-false type examination is:
- A
$\frac{7}{64}$
- ✓
$\frac{7}{128}$
- C
$\frac{45}{1024}$
- D
$\frac{7}{41}$
AnswerCorrect option: B. $\frac{7}{128}$
We know that, $\text{P}(\text{x}=\text{r})={^\text{n}}\text{C}_\text{r}(\text{P}){^\text{r}.(\text{q})6^{\text{n}-\text{r}}}$
Here, $\text{n}=10,\text{p}=\frac{1}{2},\text{q}=\frac{1}{2}$
and $\text{r}\geq8\text{ i.e,}\text{ r}=8,9,10$
$\Rightarrow\text{P}(\text{X}=\text{r})=\text{P}(\text{r}=8)+\text{P}(\text{r}=9)+\text{P}(\text{r}=10) $
$={^{10}}\text{C}_8\Big(\frac{1}{2}\Big)^8\Big(\frac{1}{2}\Big)^{10-8}+{^{10}}\text{C}_9\Big(\frac{1}{2}\Big)^9\Big(\frac{1}{2}\Big)^{10-9}+{^{10}}\text{C}_{10}\Big(\frac{1}{2}\Big)^{10}\Big(\frac{1}{2}\Big)^{10-10}$
$=\Big(\frac{10!}{8!2!}+\frac{10!}{9!1!}+1\Big)\Big(\frac{1}{2}\Big)^{10}$
$=[45+10+1]\Big(\frac{1}{2}\Big)^{10}$
$=56\Big(\frac{1}{2}\Big)^{10}=\frac{7}{128}$
View full question & answer→MCQ 351 Mark
The probability distribution of a discrete random variable $X$ is given below:
|
$\text{X}:$
|
$1$
|
$2$
|
$3$
|
$4$
|
|
$\text{P}(\text{X}):$
|
$\frac{1}{10}$
|
$\frac{1}{5}$
|
$\frac{3}{10}$
|
$\frac{2}{5}$
|
The value of $E(X^2)$ is: Answer
| $\text{X}$ |
$1$ |
$2$ |
$3$ |
$4$ |
|
| $\text{P}(\text{X})$ |
$\frac{1}{10}$ |
$\frac{1}{5}$ |
$\frac{3}{10}$ |
$\frac{2}{5}$ |
|
| $\text{X}^2\text{P(X)}$ |
$\frac{1}{10}$ |
$\frac{4}{5}$ |
$\frac{27}{10}$ |
$\frac{32}{5}$ |
$\text{E}(\text{X}^2)=10$ |
View full question & answer→MCQ 361 Mark
If a vowel is selected at random from the English alphabet then what is the probability that it is U?
- A
$\frac{1}{26}$
- ✓
$\frac{1}{5}$
- C
$\frac{5}{26}$
- D
$\frac{3}{26}$
AnswerCorrect option: B. $\frac{1}{5}$
Total number of vowels in English alphabet = 5 which are a, e, i, o, u
So, probability of u when a vowel is selected $=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac15$
View full question & answer→MCQ 371 Mark
Choose the correct answer from the given four options.
Two events E and F are independent. If $\text{P}(\text{E})=0.3,\text{P}(\text{E}\cup\text{F})=0.5,$ then $\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)-\text{P}\Big(\frac{\text{F}}{\text{E}}\Big)$ equal:
- A
$\frac{2}{7}$
- B
$\frac{3}{35}$
- ✓
$\frac{1}{70}$
- D
$\frac{1}{7}$
AnswerCorrect option: C. $\frac{1}{70}$
We have, $\text{P}(\text{E})=0.3,\text{P}(\text{E}\cup\text{F})=0.5$
Also E and F are independent.
Now $\text{P}(\text{E}\cup\text{F})=\text{P}(\text{E})+\text{P}(\text{F})-\text{P}(\text{E}\cap\text{F})$
$\Rightarrow0.5=0.3+\text{P}(\text{F})-0.3\text{P}(\text{F})$
$\Rightarrow\text{P}(\text{F})=\frac{0.5-0.3}{0.7}=\frac{2}{7}$
$\therefore\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)-\text{P}\Big(\frac{\text{F}}{\text{E}}\Big)$
$=\text{P}(\text{E})-\text{P}(\text{F})$ (as E and F are indepandent)
$=\frac{3}{10}-\frac{2}{7}=\frac{1}{70}$
View full question & answer→MCQ 381 Mark
Mark the correct alternative in the following question:
Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If $\frac{\text{P(X = r})}{\text{P(X = n} -\text{r})}$ is independent of n and r, then p equals:
- ✓
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{1}{5}$
- D
$\frac{1}{7}$
AnswerCorrect option: A. $\frac{1}{2}$
Consider,
$\text{P(X = r) = kP(X = n}-\text{r})$
Using $\text{ }^{\text{n}}\text{C}_{\text{r}}=\text{ }^{\text{n}}\text{C}_{\text{n}-\text{r}},\text{q}=1-\text{p}$
$\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}=\text{kp}^{\text{n}-\text{r}}\text{q}^{\text{r}}$
$\text{p}^{\text{r}}(1-\text{p})^{\text{n}-\text{r}}=\text{kp}^{\text{n}-\text{r}}(1-\text{p})^{\text{r}}$
$\text{p}^{2\text{r}-\text{n}}=\text{k}(1-\text{p})^{\text{2r}-\text{n}}$
$\big(\frac{\text{p}}{\text{q}}\big)^{\text{2r}-\text{n}}=\text{k}$
when p = q then k = 1
$\Rightarrow\text{p = q}=\frac{1}{2}$
View full question & answer→MCQ 391 Mark
If A and B are such that $\text{P}(\text{A}\cup\text{B})=\frac{5}{9}$ and $\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3},$ then $\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=$
- A
$\frac{9}{10}$
- ✓
$\frac{10}{9}$
- C
$\frac{8}{9}$
- D
$\frac{9}{8}$
AnswerCorrect option: B. $\frac{10}{9}$
$\text{P}(\text{A}\cup\text{B})=\frac{5}{9},\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3}$
Consider,
$\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\Rightarrow\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3}$
$\Rightarrow 1-\text{P}(\text{A}\cap\text{B})=\frac{2}{3}$
$\Rightarrow\text{P}(\text{A}\cap\text{B})=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}-\frac{5}{9}=\frac{1}{3}$
$\Rightarrow\ \text{P(A)}+\text{P(B)}=\frac{8}{9}$
$\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=1-\text{P(A)}+1-\text{P(B)}$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-\big[\text{P(A)}+\text{P(B)}\big]$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=2-\frac{8}{9}$
$\Rightarrow\text{P}(\overline{\text{A}})+\text{P}(\overline{\text{B}})=\frac{10}{9}$
View full question & answer→MCQ 401 Mark
If X follows a binomial distribution with parameter $\text{n}=100$ and $\text{p}=\frac{1}{3},$ then P(X = r) is maximum when r =
Answer$\text{n}=100,\text{p}=\frac{1}{3}\Rightarrow\text{q}=\frac{2}{3}$
$\text{np}=\frac{100}{3}=33+\frac{1}{3}$
⇒ Probability is maximum at 33.
View full question & answer→MCQ 411 Mark
Two events $A$ and $B$ will be independent, if
AnswerCorrect option: B. $\text{P}(\text{A}\ '\text{B}\ ')=\big[1-\text{P}(\text{A})\big]\big[1-\text{P}(\text{B})\big]$
Two events $A$ and $B$ are said to be independent, if $\text{P}(\text{AB})=\text{P}(\text{A})\times\text{P}(\text{B})$Distracter Rationale.
- Let $P(A) = m, P(B) = n, 0 < m, n < 1$
$A$ and $B$ are mutually exclusive.
$\therefore\text{A}\cap\text{B}=\phi$
$\Rightarrow\text{P}(\text{AB})=0$
$\text{However,}\ \text{P}(\text{A})\cdot\text{P}(\text{B})=mn\neq0$
$\therefore\text{P}(\text{A}).\text{P}(\text{B})\neq\text{P}(\text{AB})$
- Consider the result given in alternative.
$\text{P}(\text{A}'\text{B}')=\big[1-\text{P}(\text{A})\big]\big[1-\text{P}(\text{B})\big]$
$\Rightarrow\text{P}(\text{A}'\cap\text{B}')=1-\text{P}(\text{A})-\text{P}(\text{B})+\text{P}(\text{A}).\text{P}(\text{B})$
$\Rightarrow1-\text{P}(\text{A}\cup\text{B})=1-\text{P}(\text{A})-\text{P}(\text{B})+\text{P}(\text{A}).\text{P}(\text{B})$
$\Rightarrow\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A})\cdot\text{P}(\text{B})$
$ \Rightarrow\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{AB})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}).\text{P}(\text{B})$
$\Rightarrow\text{P}(\text{AB})=\text{P}(\text{A}).\text{P}(\text{B})$
This implies that $A$ and $B$ are independent, if $\text{P}(\text{A}\ '\text{B}\ ')=\big[1-\text{P}(\text{A})\big]\big[1-\text{P}(\text{B})\big]$
- Let $A:$ Event of getting an odd number on throw of a die $= \{1, 3, 5\}$
$\Rightarrow\text{P}(\text{A})=\frac{3}{6}=\frac{1}{2}$
$B:$ Event of getting an even number on throw of a die $= \{2, 4, 6\}$
$\text{P}(\text{B})=\frac{3}{6}=\frac{1}{2}$
Here, $\text{A}\cap\text{B}=\phi$
$\therefore\text{P}(\text{AB})=0 $
$\text{P}(\text{A}).\text{P}(\text{B})=\frac{1}{4}\neq0$
$ \Rightarrow\text{P}(\text{A}).\text{P}(\text{B})\neq\text{P}(\text{AB})$
- From the above example, it can be seen that,
$\text{P}(\text{A})+\text{P}(\text{B})=\frac{1}{2}+\frac{1}{2}=1$
However, it cannot be inferred that $A$ and $B$ are independent.
Thus, the correct answer is $B.$ View full question & answer→MCQ 421 Mark
Choose the correct answer from the given four options.If two events are independent, then:
- A
They must be mutually exclusive.
- B
The sum of their probabilities must be equal to 1.
- C
(a) and (b) both are correct.
- ✓
None of the above is correct.
AnswerCorrect option: D. None of the above is correct.
If two events A and B are independent, then we know that
$\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})\cdot\text{P}(\text{B}),\text{P}(\text{A})\neq0,\text{P}(\text{B})\neq0$
Since, A and B have a common outcome.
Further, mutually exclusive events never have a common outcome.
In other words, two independents events having non-zero probabilities of occurrence cannot be mutually exclusive and conversely, i.e., two mutually exclusive events having non-zero probabilities of outcome cannot be independent.
View full question & answer→MCQ 431 Mark
The probability distribution of a discrete random variable X is given below:
| $\text{X}:$ |
$2$ |
$3$ |
$4$ |
$5$ |
| $\text{P}(\text{X}):$ |
$\frac{5}{\text{k}}$ |
$\frac{7}{\text{k}}$ |
$\frac{9}{\text{k}}$ |
$\frac{11}{\text{k}}$ |
The value of k is: Answer$\sum\limits_2^5\text{P}(\text{x})=1$
$\frac{5}{\text{k}}+\frac{7}{\text{k}}+\frac{9}{\text{k}}+\frac{11}{\text{k}}=1$
$\text{k}=32$
NOTE: Question is modified.
View full question & answer→MCQ 441 Mark
If A and B are independent events such that P(A) > 0 and P(B) > 0, then.
- A
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}).\text{P}(\text{B})$
- ✓
$\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A}).\text{P}(\text{B})$
- C
$\text{P}\big(\frac{\text{A}}{\text{B}}\big)=\text{P}(\text{A})$
- D
$\text{P}\big(\frac{\text{B}}{\text{A}}\big)=\text{P}(\text{B})$
AnswerCorrect option: B. $\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A}).\text{P}(\text{B})$
Since, A and B are independent events
$\therefore\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A}).\text{P}(\text{B})$
$\text{P}\big(\frac{\text{A}}{\text{B}}\big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\text{P}(\text{A})$
and $\text{P}\big(\frac{\text{B}}{\text{A}}\big)=\frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}=\text{P}(\text{B})$
View full question & answer→MCQ 451 Mark
A letter is known to have come either from $\text{LONDON}$ or $\text{CLIFTON};$ on the postmark only the two consecutive letters $ON$ are ellegible. The probability that it came from $\text{LONDON}$ is:
- A
$\frac{5}{17}$
- ✓
$\frac{12}{17}$
- C
$\frac{17}{30}$
- D
$\frac{3}{5}$
AnswerCorrect option: B. $\frac{12}{17}$
We define the following events:
$A_1:$ Selecting a pair of consecutive letters from the word $\text{LONDON}$
$A_2:$ Selecting a pair of consecutive letters from the word $\text{CLIFTON}$
$E:$ Selecting a pair of letters $ON$
Then ${\text{P(A}_1∩\text{E})=\frac52},$ as there are $5$ pairs of consecutive letters out of which $2$ are $ON.$
$\text{P(A}_2∩\text{E})=\frac61,$ as there are $6$ pairs of consecutive letters of which $1$ is $ON.$
So, required probability $\text{P}=\Big(\frac{\text{A}_1}{\text{E}}\Big)$
$\Rightarrow\Big(\frac{\text{A}_1}{\text{E}}\Big)=\frac{\text{P}(\text{A}_1\cap\text{E})}{\text{P}(\text{A}_1\cap\text{E}) + \text{P}(\text{A}_1\cap\text{E})}=\frac{\frac25}{\frac25+\frac16}=\frac{12}{17}$
View full question & answer→MCQ 461 Mark
Choose the correct answer from the given four options.If $\text{P}(\text{A})=\frac{2}{5},\text{P}(\text{B})=\frac{3}{5}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{5},$ then $\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)\cdot\text{P}\Big(\frac{\text{B}'}{\text{A}'}\Big)$ is equas:
- ✓
$\frac{5}{6}$
- B
$\frac{5}{7}$
- C
$\frac{25}{42}$
- D
$1$
AnswerCorrect option: A. $\frac{5}{6}$
Here, $\text{P}(\text{A})=\frac{2}{5},\text{P}(\text{B})=\frac{3}{5}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{5}$
$\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)=\frac{\text{P}(\text{A}'\cap\text{B}')}{\text{P}(\text{B}')}+\frac{1-\text{P}(\text{A}\cap\text{B})}{1-\text{P}(\text{B})}$
$=\frac{1-\big[\text{P}(\text{A})+\text{P}(\text{B})-\text{A}(\text{A}\cap\text{B})\big]}{1-\text{P}(\text{B})}$
$=\frac{1-\Big(\frac{2}{5}+\frac{3}{10}-\frac{1}{5}\Big)}{1-\frac{3}{10}}$
$=\frac{1-\Big(\frac{4+3-2}{10}\Big)}{\frac{7}{10}}-\frac{1-\frac{1}{2}}{\frac{7}{10}}$
$=\frac{5}{7}$
And $\text{P}\Big(\frac{\text{B}'}{\text{A}'}\Big)=\frac{\text{P}(\text{B}'\cap\text{A}')}{\text{P}(\text{A}')}$
$=\frac{1-\text{P}(\text{A}\cup\text{B})}{1-\text{P}(\text{A})}$
$=\frac{1-\frac{1}{2}}{1-\frac{2}{5}}$ $\Big[\because\text{P}(\text{A}\cup\text{B})=\frac{1}{2}\Big]$
$=\frac{\frac{1}{2}}{\frac{3}{5}}=\frac{5}{6}$
$\therefore\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)\cdot\text{P}\Big(\frac{\text{B}'}{\text{A}'}\Big)=\frac{5}{7}\cdot\frac{5}{6}=\frac{25}{42}$
View full question & answer→MCQ 471 Mark
A bag $A$ contains $4$ green and $6$ red balls. Another bag $B$ contains $3$ green and $4$ red balls. If one ball is drawn from each bag, find the probability that both are green:
- A
$\frac{13}{70}$
- B
$\frac{1}{4}$
- ✓
$\frac{6}{35}$
- D
$\frac{8}{35}$
AnswerCorrect option: C. $\frac{6}{35}$
Bag $A$ has $4$ green balls and $6$ red balls
$\Rightarrow$ probability of choosing green ball from $A$ is $p(\text{green}_A) = \frac4{10}$
Bag $A$ has $3$ green balls and $4$ red balls
$\Rightarrow$ probability of choosing green ball from $B$ is $p(\text{green}_B) =\frac37$
On choosing one ball from each bag probability that both are green $= p(\text{green}_A) \times p(\text{green}_B)$
$=\frac4{10}\times\frac37=\frac{6}{35}$
View full question & answer→MCQ 481 Mark
In a college 30% students fail in Physics, 25% fail in Mathenatics and 10% fail in both. One student is chosen at random. The probability that she fails in Physics if she failed in Mathematics is.
- A
$\frac{1}{10}$
- B
$\frac{1}{3}$
- ✓
$\frac{2}{5}$
- D
$\frac{9}{20}$
AnswerCorrect option: C. $\frac{2}{5}$
Let A be the event that students failed in Physics. B be the event that students failed in Mathematics.
Given that, $\text{P(A)}=30\%=\frac{30}{100}$
$\text{P(B)}=25\%=\frac{25}{100}$
$\text{P}(\text{A}\cap\text{B})=10\%=\frac{10}{100}$
Required probability is given by $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{\frac{10}{100}}{\frac{25}{100}}=\frac{2}{5}$
View full question & answer→MCQ 491 Mark
If $\text{P(B)}=\frac{3}{5},\text{P}(\text{A}|\text{B})=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $\text{P}(\overline{\text{A}\cap\text{B}})+\text{P}(\overline{\text{A}}\cap\text{B})=$
- A
$\frac{1}{5}$
- B
$\frac{4}{5}$
- C
$\frac{1}{2}$
- ✓
$1$
Answer$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}\Big({\text{A}}\cup{\text{B}}\Big)=\frac{4}{5}$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{2}$
$\Rightarrow\ \frac{\text{P}(\text{A}\cap\text{B})}{\frac{3}{5}}=\frac{1}{2}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{3}{10}$
$\text{P}(\overline{\text{A}\cup\text{B}})+\text{P}(\overline{\text{A}}\cap\text{B})$
$=1-\text{P}(\text{A}\cap\text{B})+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=1-\frac{3}{10}+\frac{3}{5}-\frac{3}{10}$
$=1$
View full question & answer→MCQ 501 Mark
Let X be a discrete random variable. Then the variance of X is:
AnswerCorrect option: C. $E(X^2) - (E(X))^2$
Since, the variance of a discrete random variable X is given by:
Var$(X) = E(X^2) - (E(X))^2$
Hence, the correct alternative is option (c).
View full question & answer→