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
$\frac{1}{3}$
- C
$\frac{3}{4}$
- D
$1$
Answer - $\frac{3}{4}$
Solution:
$\text{P(A)}=\frac{1}{3},\text{ P(B)}=\frac{1}{4},\text{ P}\Big(\frac{\text{B}'}{\text{A}}\Big)=?$
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P(B)}$
$\text{P}\Big(\frac{\text{B}'}{\text{A}}\Big)=\text{P}\frac{(\text{A}\cap\text{B}')}{\text{P(A)}}$
$\text{P}(\text{B}')=1-\frac{1}{4}=\frac{3}{4}$
$\therefore\text{P}\Big(\frac{\text{B}'}{\text{A}}\Big)=\frac{\text{P(A)}\times\text{P(B})'}{\text{P(A)}}$
$=\text{P(B}')=\frac{3}{4}$
View full question & answer→MCQ 21 Mark
The probability distribution of a random variable $X$ is:| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 0.1 | k | 2k | k | 0.1 |
where $k$ is some unknown constant.
The probability that the random variable $X$ takes the value 2 is: - A
$\frac{1}{5}$
- B
$\frac{2}{5}$
- C
$\frac{4}{5}$
- D
Answer$\begin{array}{l}\text {As } \sum_{i=1}^n p_i=1 \Rightarrow 0.1+k+2 k+k+0.1=1 \\ \Rightarrow \quad 4 k=1-0.2=0.8 \Rightarrow k=\frac{0.8}{4}=0.2 \\ \therefore \quad P(x=2)=2 k=2 \times 0.2=0.4=\frac{4}{10}=\frac{2}{5}\end{array}$
View full question & answer→MCQ 31 Mark
Let $E$ be an event of a sample space $S$ of an experiment then $P(S \mid E)=$
- A
$P(S \cap E)$
- B
$P(E)$
- C
- D
$0$
Answer$P(S \mid E)=\frac{P(S \cap E)}{P(E)}=\frac{P(E)}{P(E)}=1$
View full question & answer→MCQ 41 Mark
If $A$ and $B$ are events such that $P(A / B)=P(B / A) \neq 0$, then
AnswerGiven, $P(A / B)=P(B / A)$
$
\begin{array}{l}
\Rightarrow \quad \frac{P(A \cap B)}{P(B)}=\frac{P(B \cap A)}{P(A)} \\
\Rightarrow \quad P(A)=P(A) \quad[\because P(A \cap B)=P(B \cap A)]
\end{array}$
View full question & answer→MCQ 51 Mark
A problem in Mathematics is given to three students whose chances of solving it are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ respectively. If the events of their solving the problem are independent then the probability that the problem will be solved, is
- A
$\frac{1}{4}$
- B
$\frac{1}{3}$
- C
$\frac{1}{2}$
- D
$\frac{3}{4}$
AnswerLet $A, B, C$ be the respective events of solving the problem. Then, $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(C)=\frac{1}{4}$. Here, $A, B$, $C$ are independent events. Problem is solved if at least one of them solves the problem.
Required probability is $=P(A \cup B \cup C)=1-P(\bar{A}) P(\bar{B}) P(\bar{C})$
$
=1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)=1-\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4}=1-\frac{1}{4}=\frac{3}{4} \text {. }$
View full question & answer→MCQ 61 Mark
$M$ and $N$ are two events such that $P(M \cap N)=0$. Which of the following is equal to $P(M \mid(M \cup N))$ ?
AnswerCorrect option: C. $\frac{P(M)}{P(M)+P(N)}$
$\frac{P(M)}{P(M)+P(N)}$
View full question & answer→MCQ 71 Mark
Five fair coins are tossed simultaneously. The probability of the events that atleast one head comes up is
- A
$\frac{27}{32}$
- B
$\frac{5}{32}$
- C
$\frac{31}{32}$
- D
$\frac{1}{32}$
AnswerSince each coin turns up on either a head or tail.
$\therefore \quad$ Total possible outcomes $=2^5=32$
Let $A$ be the event that all tails comes up.
$\therefore \quad n(A)=1\{$ i.e., $(T, T, T, T, T)$
So, required probability $=1-P(A)=1-\frac{1}{32}=\frac{31}{32}$
View full question & answer→MCQ 81 Mark
For two events $A$ and $B$, if $P(A)=0.4, P(B)=0.8$ and $P(B / A)=0.6$, then $P(A \cup B)$ is
- A
$0.24$
- B
$0.3$
- C
$0.48$
- D
$0.96$
AnswerWe have, $P(A)=0.4, P(B)=0.8$ and $P(B / A)=0.6$
We know that $P(B / A)=\frac{P(A \cap B)}{P(A)}$
$
\Rightarrow 0.6=\frac{P(A \cap B)}{0.4} \Rightarrow P(A \cap B)=0.24
$
Now: $P(A \cup B)=P(A)+P(B)-P(A \cap B)=0.4+0.8-0.24=0.96$
Hence. $P(A \cup B)=0.96$
View full question & answer→MCQ 91 Mark
If $P\left(\frac{A}{B}\right)=0.3, P(A)=0.4$ and $P(B)=0.8$, then $P\left(\frac{B}{A}\right)$ is equal to:
- A
$0.6$
- B
$0.3$
- C
$0.06$
- D
$0.4$
AnswerWe have, $P\left(\frac{A}{B}\right)=0.3, P(A)=0.4$ and $P(B)=0.8$
Now, $P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}$
$
\Rightarrow \quad 0.3=\frac{P(A \cap B)}{0.8} \Rightarrow P(A \cap B)=0.3 \times 0.8=0.24
$
Now, $P\left(\frac{B}{A}\right)=\frac{P(B \cap A)}{P(A)}=\frac{0.24}{0.4}=0.6$
View full question & answer→MCQ 101 Mark
A family has 2 children and the elder child is a girl. The probability that both children are girls is:
- A
$\frac{1}{4}$
- B
$\frac{1}{8}$
- C
$\frac{1}{2}$
- D
$\frac{3}{4}$
AnswerLet $G_i(i=1,2)$ and $B_i(i=1,2)$ denote the $i^{\text {th }}$ child is a girl or a boy respectively.
Then sample space, $S=\left\{G_1 G_2, G_1 B_2, B_1 G_2, B_1 B_2\right\}$
Let $A$ be the event that both children are girls,
$B$ be the event that the elder child is a girl.
Thus, $A=\left\{G_1 G_2\right\}$ and $B=\left\{G_1 G_2, G_1 B_2\right\}$
$\Rightarrow A \cap B=\left\{G_1 G_2\right\}$
$\therefore \quad$ Required probability $=P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{1 / 4}{2 / 4}=\frac{1}{2}$
View full question & answer→MCQ 111 Mark
If for any two events $A$ and $B$, $P(A)=\frac{4}{5}$ and $P(A \cap B)=\frac{7}{10}$, then $P(B / A)$ is
- A
$\frac{1}{10}$
- B
$\frac{1}{8}$
- C
$\frac{7}{8}$
- D
$\frac{17}{20}$
AnswerWe know that, $P(B / A)=\frac{P(B \cap A)}{P(A)}=\frac{7 / 10}{4 / 5}=\frac{7}{8}$
View full question & answer→MCQ 121 Mark
A card is picked at random from a pack of 52 playing cards. Given that the picked card is a queen, the probability of this card to be a card of spade is
- A
$\frac{1}{3}$
- B
$\frac{4}{13}$
- C
$\frac{1}{4}$
- D
$\frac{1}{2}$
AnswerLet $A$ be the event that the card is a spade and $B$ be the event that the picked card is a queen.
We have a total of 13 spades and 4 queen cards.
Also only one queen is from spade.
$
\therefore \quad P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{52}}{\frac{4}{52}}=\frac{1}{4}
$
View full question & answer→MCQ 131 Mark
A die is thrown once. Let $A$ be the event that the number obtained is greater than 3 . Let $B$ be the event that the number obtained is less than 5 . Then $P(A \cup B)$ is
- A
$\frac{2}{5}$
- B
$\frac{3}{5}$
- C
$0$
- D
Answer$\begin{array}{l}\text { } \text { Here, } A=\{4,5,6\}, B=\{1,2,3,4\} \\ A \cap B=\{4\} \\ \text { Now, } P(A \cup B)=P(A)+P(B)-P(A \cap B)=\frac{3}{6}+\frac{4}{6}-\frac{1}{6}=1\end{array}$
View full question & answer→MCQ 141 Mark
If $A$ and $B$ are two independent events with $P(A)=\frac{1}{3}$ and $P(B)=\frac{1}{4}$, then $P\left(B^{\prime} \mid A\right)$ is equal to
- A
$\frac{1}{4}$
- B
$\frac{1}{3}$
- C
$\frac{3}{4}$
- D
AnswerGiven, $A$ and $B$ are independent events.
Also, $P(A)=\frac{1}{3}$ and $P(B)=\frac{1}{4}$
$
\begin{array}{l}
\text { Now, } P\left(B^{\prime} \mid A\right)=\frac{P\left(B^{\prime} \cap A\right)}{P(A)} \\
=\frac{P\left(B^{\prime}\right) P(A)}{P(A)} \quad[\because A, B \text { are independent events }] \\
=P\left(B^{\prime}\right)=1-P(B)=1-\frac{1}{4}=\frac{3}{4}
\end{array}$
View full question & answer→MCQ 151 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
$\frac{18}{108}$
- C
$\frac{30}{108}$
- D
$\frac{48}{108}$
Answer - $\frac{48}{108}$
Solution:
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 161 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 171 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
$\frac{3}{35}$
- C
$\frac{1}{70}$
- D
$\frac{1}{7}$
Answer - $\frac{1}{70}$
Solution:
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 181 Mark
A rifleman is firing at a distant target and has only 10% chance of hiting it. the least number of round he must fire in order to have more than 50% chance of hitting it at least once is:
Answer - 7
Solution:
Given $\text{p}=\frac{1}{10}\Rightarrow\text{q}=\frac{9}{10}$
Let n be the number of rounds.
$\text{P(X}\geq1)=1-\text{P(X}=0)$
$\Rightarrow\text{P(X}\geq1)\geq0.5$
$\Rightarrow1-\text{P(X}=0)\geq0.5$
$\Rightarrow\text{P(X}=0)\leq0.5$
$\Rightarrow0.9^{\text{n}}\leq0.5$
Using log table,
$\text{n}\leq6.572\approx7$
He must fire in order to have more than
50% chance of hitting the target at least once.
View full question & answer→MCQ 191 Mark
A and B draw two cards each, one after another, from a pack of well-shuffled pack of 52 cards. The probability that all the four cards drawn are of the same suit is
Answer - $\frac{44}{85\times49}$
Solution:
Total cards = 52 There are four suits of cards in a pack, i.e. diamond, heart, spade and club.
Pall 4 cards are of same suit = Pall 4 cards are of diamond + Pall 4 cards are of heart + Pall 4 cards are of spade + Pall 4 cards are of club.
$=4\times\frac{13}{52}\times\frac{12}{51}\times\frac{11}{50}\times\frac{10}{59}$
$=4\times\frac{11}{85\times49}$
$=\frac{44}{85\times49}$
View full question & answer→MCQ 201 Mark
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?
Answer - 0, 1, 2
Solution:
The two balls selected can be represented as BB, BR, RB, RR, where B represents a black ball and R represents a red ball.
X represents the number of black balls.
$\therefore$ X(BB) = 2
X(BR) = 1
X(RB) = 1
X(RR) = 0
Therefore, the possible values of X are 0, 1 and 2.
Yes, X is a random variable.
View full question & answer→MCQ 211 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
$\frac{1}{10}$
- B
$\frac{2}{5}$
- C
$\frac{9}{20}$
- D
$\frac{1}{3}$
Answer - $\frac{2}{5}$
Solution:
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 221 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(X2) is equal to:
Answer - 10.
solution:
$\text{E}(\text{X}^2)=\sum\text{X}^2\text{P}(\text{X})$
$=1\cdot\frac{1}{10}+4\cdot\frac{1}{5}+9\cdot\frac{3}{10}+16\cdot\frac{2}{5}$
$=\frac{1}{10}+\frac{4}{5}+\frac{27}{10}+\frac{32}{5}$
$=\frac{1+8+27+64}{10}=10$
View full question & answer→MCQ 231 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
$\frac{1}{3}$
- D
$\frac{2}{9}$
Answer - $\frac{2}{9}$
Solution:
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 241 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 - -1.8
Solution:
$\text{E}(\text{X})=\sum\text{P}(\text{X})$
$= -4\times(0.1)+(-3 \times0.2)+(-2\times0.3)+(-1\times0.2)+(0\times0.2)$
$=-0.4-0.6-0.6-0.2=-1.8$
View full question & answer→MCQ 251 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,
- A
$\frac{5}{84}$
- B
$\frac{3}{9}$
- C
$\frac{3}{7}$
- D
$\frac{7}{17}$
Answer - $\frac{5}{84}$
Solution:
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 261 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
$\frac{1}{5}$
- B
$\frac{2}{5}$
- C
$\frac{3}{5}$
- D
$\frac{4}{5}$
Answer - $\frac{2}{5}$
Solution:
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 271 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:
- A
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{1}{5}$
- D
$\frac{1}{7}$
Answer - $\frac{1}{2}$
Solution:
$\text{P}(\text{x}=\text{r})={^\text{n}}\text{C}_\text{r}(\text{p})^\text{r}(\text{q})^{\text{n}-\text{r}}$
$=\frac{\text{n}!}{(\text{n}-\text{r})!\text{r}!}(\text{p})^\text{r}(1-\text{p})^{\text{n}-\text{r}}[\therefore\text{q}=1-\text{p}]$
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 281 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
$\frac{3}{20}$
- D
$\frac{6}{25}$
Answer - $\frac{6}{25}$
Solution:
$\text{P(A)}=\frac{3}{8},\text{P(B)}=\frac{5}{8},\text{P}\Big({\text{A}}\cup{\text{B}}\Big)=\frac{3}{4}$
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \frac{3}{4}=\frac{3}{8}+\frac{5}{8}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{4}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}\times\frac{\text{P}(\overline{\text{A}}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}\times\frac{\text{P(B)}-\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{\frac{1}{4}}{\frac{5}{8}}\times\frac{\big(\frac{5}{8}-\frac{1}{4}\big)}{\frac{5}{8}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\text{ P}\Big(\frac{\overline{\text{A}}}{\text{B}}\Big)=\frac{6}{25}$
View full question & answer→MCQ 291 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
$\frac{7}{8}$
- D
$\frac{15}{16}$
Answer - $\frac{15}{16}$
Solution:
$\text{E(X)}=2, \text{V(X})=1$
$\text{np}=2,\text{npq}=1$
$\Rightarrow\text{q}=\frac{1}2{}=\text{p}$
$\Rightarrow\text{n}=4$
$\text{P(X}\geq1)=1-\text{P(X}=0)$
$\text{P(X}\geq1)=1-\big(\frac{1}{2}\big)^4$
$\text{P(X}\geq1)=1-\frac{1}{16}=\frac{15}{16}$
View full question & answer→MCQ 301 Mark
A die is thrown and a card is selected ar random from a deck pf 52 playing cards. The probability of getting an even number of the die and a spade card is
- A
$\frac{1}{2}$
- B
$\frac{1}{4}$
- C
$\frac{1}{8}$
- D
$\frac{3}{4}$
Answer - $\frac{1}{8}$
Solution:
A Sample space when a die is thrown,
S1 = {1, 2, 3, 4, 5, 6} ⇒ n(S1) = 6
Let A be the event that getting even number.
A = {2, 4, 6} ⇒ 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) × P(B)
$=\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}$
View full question & answer→MCQ 311 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,
- A
$\frac{^{7}\text{P}_5}{7_5}$
- B
$\frac{7_5}{^{7}\text{P}_5}$
- C
$\frac{6}{^{6}\text{P}_5}$
- D
$\frac{^{5}\text{P}_5}{5}$
Answer - $\frac{^{7}\text{P}_5}{7_5}$
Solution:
Five persons can leave different floors
By 7P5 ways.
Possible ways of leavinf the lift = 75
Required probability $=\frac{^{7}\text{P}_5}{7^5}$
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.
- A
$\frac{10}{13}$
- B
$\frac{13}{120}$
- C
$\frac{1}{40}$
- D
$\frac{1}{12}$
Answer - $\frac{10}{13}$
Solution:
Let E1 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 E2 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
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
$\frac{7}{64}$
- B
$\frac{7}{128}$
- C
$\frac{45}{1024}$
- D
$\frac{7}{41}$
Answer - $\frac{7}{128}$
Solution:
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 341 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
Answer - 0.188
Solution:
Let:
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 351 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
$\frac{9}{10}$
- B
$\frac{10}{9}$
- C
$\frac{8}{9}$
- D
$\frac{9}{8}$
Answer - $\frac{10}{9}$
Solution:
$\text{P}(\text{A}\cup\text{B})=\frac{5}{9},\text{P}(\overline{\text{A}}\cup\overline{\text{B}})=\frac{2}{3}$
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 361 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 - 33
Solution:
$\text{n}=100,\text{p}=\frac{1}{3}\Rightarrow\text{q}=\frac{2}{3}$
$\text{np}=\frac{100}{3}=33+\frac{1}{3}$
⇒ Probability is maximum at 33.
View full question & answer→MCQ 371 Mark
If the random variable X has the following distribution:
| X: | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X): | a | 3a | 5a | 7a | 9a | 11a | 13a | 15a | 17a |
then the value of a is:
- A
$\frac{7}{81}$
- B
$\frac{5}{81}$
- C
$\frac{2}{81}$
- D
$\frac{1}{81}$
Answer - $\frac{1}{81}$
Solution:
We know that the sum of probsabilities in a probability distribution is always 1.
$\therefore$ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 1
⇒ a + 3a+ 5a+ 7a+ 9a + 11a + 13a + 15a + 17a = 1
⇒ 81a = 1
$\Rightarrow\text{a}=\frac{1}{81}$
View full question & answer→MCQ 381 Mark
A biased coin with probabilty p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $\frac{2}{5},$ then p equals:
- A
$\frac{1}{3}$
- B
$\frac{2}{3}$
- C
$\frac{2}{5}$
- D
$\frac{3}{5}$
Answer - $\frac{1}{3}$
Solution:
p is the probability of getting head.
q = 1 - p is the probability of getting tail.
The number of tosses required is even.
$\Rightarrow\text{qp+q}^3\text{p+q}^5\text{p+q}^7\text{p+q}^9\text{p}\dots$
$\Rightarrow\text{qp}\Big(\frac{1}{1-\text{q}^2}\Big)$
$\Rightarrow\frac{(1-\text{p})\text{p}}{1-(1-\text{p})^2}$
$\Rightarrow\frac{(1-\text{p})\text{p}}{1-(1-2\text{p + p}^2)}$
$\Rightarrow\frac{1-\text{p}}{2-\text{p}}$
Given $\frac{1-\text{p}}{2-\text{p}}=\frac{2}{5}$
$\Rightarrow\text{p}=\frac{1}{3}$
View full question & answer→MCQ 391 Mark
If A and B are two events such that $\text{P}(\text{A}|\text{B})=\text{p},\text{P(A)}=\text{p},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cup\text{B})=\frac{5}{9},$ then p =
- A
$\frac{2}{3}$
- B
$\frac{3}{5}$
- C
$\frac{1}{3}$
- D
$\frac{3}{4}$
Answer - $\frac{1}{3}$
Solution:
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{p},\text{P(A)}=\text{p},\text{P(B)}=\frac{1}{3},\text{P}(\text{A}\cup\text{B})=\frac{5}{9}$
Consider,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{p}$
$\Rightarrow \frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\text{P}$
$\Rightarrow \frac{\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})}{\text{P(B)}}=\text{P}$
$\Rightarrow\frac{\text{p}+\frac{1}{3}-\frac{5}{9}}{\frac{1}{3}}=\text{p}$
$\Rightarrow\text{p}+\frac{1}{3}-\frac{5}{9}=\frac{\text{p}}{3}$
$\Rightarrow\frac{-2}{9}=\frac{\text{p}}{3}-\text{p}$
$=\frac{-2}{3}\text{p}=\frac{-2}{9}$
$\Rightarrow\text{p}=\frac{1}{3}$
View full question & answer→MCQ 401 Mark
A pack of playing cards was found to contain only 51 cards. If the first 13 cards which are examined are all red, then the probability thatthe missing card is black, is:
Answer - $\frac23$
Solution:
Total number of cards = 52
Number of lost cards = 1
13 cards are surley red therfore, from the remaining 39 cards 26 are black and 13 are red.
So probabilityof lost card being black $=\frac{(261)}{(391)}=\frac{26}{39}=\frac{2}{3}$
View full question & answer→MCQ 411 Mark
Choose the correct answer from the given four options:
let $\text{P}(\text{A})=\frac{7}{13},\text{P}(\text{B})=\frac{9}{13}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{13}.$ Then $\text{P}\Big(\frac{\text{A'}}{\text{B}}\Big)$ is equal to:
- A
$\frac{6}{13}$
- B
$\frac{4}{13}$
- C
$\frac{4}{9}$
- D
$\frac{5}{9}$
Answer - $\frac{5}{9}$
Solution:
Here, $\text{P}(\text{A})=\frac{7}{13},\text{P}(\text{B})=\frac{9}{13}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{13}$
$\therefore\text{P}\Big(\frac{\text{A}'}{\text{B}}\Big)=\frac{\text{P}(\text{A}'\cap\text{B})}{\text{P}(\text{B})}=\frac{\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$=\frac{\frac{9}{13}-\frac{4}{13}}{\frac{9}{13}}=\frac{\frac{5}{13}}{\frac{9}{13}}=\frac{5}{9}$
View full question & answer→MCQ 421 Mark
Associated to a random experiment two events A and B are such that $\text{P(B)}=\frac{3}{5},\text{P}(\text{A}|\text{B})=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$. The value pf P(A) is
- A
$\frac{3}{10}$
- B
$\frac{1}{2}$
- C
$\frac{1}{10}$
- D
$\frac{3}{5}$
Answer - $\frac{1}{2}$
Solution:
$\text{P(B)}=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2},\text{P}(\text{A}\cup\text{B})=\frac{4}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{2}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\frac{3}{5}}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{3}{10}$
$\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=\frac{3}{10}$
$\text{P(A)}+\frac{3}{5}-\frac{4}{5}=\frac{3}{10}$
$\text{P(A)}=\frac{1}{2}$
View full question & answer→MCQ 431 Mark
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?
Answer - 0, 2, 4, 6
Solution:
A coin is tossed six times and X represents the difference between the number of heads and the number of tails.
$\therefore$ X(6H, 0T)=∣6 - 0∣ = 6
X(5H, 1T) = ∣5 - 1∣ = 4
X(4H, 2T) = ∣4 - 2∣ = 2
X(3H, 3T) = ∣3 - 3∣ = 0
X(2H, 4T) = ∣2 - 4∣ = 2
X(1H, 5T) = ∣1 - 5∣ = 4
X(0H, 6T) = ∣0 - 6∣ = 6
Thus, the possible values of X are 0, 2, 4 and 6.
View full question & answer→MCQ 441 Mark
If A and B are two events such that $\text{P(A)}=\frac{1}{2},\text{P(B)}=\frac{1}{3},\text{P}(\text{A}|\text{B})=\frac{1}{4},$ then $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$ equals.
- A
$\frac{1}{12}$
- B
$\frac{3}{4}$
- C
$\frac{1}{4}$
- D
$\frac{3}{16}$
Answer - $\frac{1}{4}$
Solution:
$\text{P(A)}=\frac{1}{2},\text{P(B)}=\frac{1}{3},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{4}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{4}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{1}{4}$
$\frac{\text{P}(\text{A}\cap\text{B})}{\frac{1}{3}}=\frac{1}{4}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{12}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{1}{12}}{\frac{1}{3}}=\frac{1}{4}$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\text{P}(\text{A}\cup\text{B})$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=-1\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\Big[\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\Big]$
$\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\frac{1}{4}$
View full question & answer→MCQ 451 Mark
Choose the correct answer from the given four options.
A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2 green balls and one blue ball is:
- A
$\frac{3}{28}$
- B
$\frac{2}{21}$
- C
$\frac{1}{28}$
- D
$\frac{167}{168}$
Answer - $\frac{3}{28}$
Solution:
Probability of drawing 2 green balls and one blue ball
$=\text{P}_\text{G}\cdot\text{P}_\text{G}\cdot\text{P}_\text{B}+\text{P}_\text{B}\cdot\text{P}_\text{G}\cdot\text{P}_\text{G}+\text{P}_\text{G}\cdot\text{P}_\text{B}\cdot\text{P}_\text{G}$
$=\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{2}{6}+\frac{2}{8}\cdot\frac{3}{7}\cdot\frac{2}{6}+\frac{3}{8}\cdot\frac{2}{7}\cdot\frac{2}{6}$
$=\frac{1}{28}+\frac{1}{28}+\frac{1}{28}=\frac{3}{28}$
View full question & answer→MCQ 461 Mark
If A and B are two events such that $\text{P(A)}=\frac{4}{5},$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{10},$ then P(B|A) =
- A
$\frac{1}{10}$
- B
$\frac{1}{8}$
- C
$\frac{7}{8}$
- D
$\frac{17}{20}$
Answer - $\frac{7}{8}$
Solution:
We have,
$\text{P(A)}=\frac{4}{5}$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{10}$
Now,
$\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{\Big(\frac{7}{10}\Big)}{\Big(\frac{4}{5}\Big)}$
$=\frac{7\times5}{10\times4}$
$=\frac{7}{8}$
Hence, the correct alternative is option (c).
View full question & answer→MCQ 471 Mark
If the events A and B are independent, then $\text{P}(\text{A}\cap\text{B})$ is equal to,
Answer - P(A) P(B)
Solution:
$\text{P}(\text{A}\cap\text{B})=\text{P(A)} \text{ P(B)}$ for independent events.
View full question & answer→MCQ 481 Mark
A box contains 10 good articles and 6 with defects. One item is drawn at random. The probability that it is either good or has a defect is,
- A
$\frac{64}{64}$
- B
$\frac{49}{64}$
- C
$\frac{40}{64}$
- D
$\frac{24}{64}$
Answer - $\frac{64}{64}$
Solution:
P(good item) $=\frac{10}{16}$
P(defected item) $=\frac{6}{16}$
P(eitherr good or defected item) = P(good item) + P(defected item)
$=\frac{10}{16}+\frac{6}{16}$
$=\frac{16}{16}$
$=1$
$=\frac{64}{64}$
View full question & answer→MCQ 491 Mark
If A and B are two independent events such that P(A) = 0.3 and $\text{P}(\text{A}\cup\text{B})=0.5,$ then P(A|B) - P(B|A) =
- A
$\frac{2}{7}$
- B
$\frac{3}{35}$
- C
$\frac{1}{70}$
- D
$\frac{1}{7}$
Answer - $\frac{1}{70}$
Solution:
We have,
$\text{P(A)}=0.3$ and $\text{P}(\text{A}\cup\text{B})=0.5$
As, A and B are independent events
So, $\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P(B)}$
$=0.3\times\text{P(B)}$
$=0.3\text{ P(B)}\ .....\text{(i)}$
Also, $\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow 0.5 = 0.3+\text{P(B)}-0.3\text{ P(B)}$ [Using (i)]
$\Rightarrow 0.5-0.3 = 0.7\text{ P(B)}$
$\Rightarrow0.7\text{ P(B)}=0.2$
$\Rightarrow\text{ P(B)}=\frac{0.2}{0.7}$
$\Rightarrow\text{ P(B)}=\frac{2}{7}$
Using (i), we get
$\text{P}(\text{A}\cap\text{B})=0.3\times\frac{2}{7}=\frac{6}{70}$
Now,
$\text{P}(\text{A}|\text{B})-\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}-\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{\Big(\frac{6}{70}\Big)}{\Big(\frac{2}{7}\Big)}-\frac{\Big(\frac{6}{70}\Big)}{0.3}$
$=\frac{6\times7}{70\times2}-\frac{6}{70\times0.3}$
$=\frac{3}{10}-\frac{2}{7}$
$=\frac{21-20}{70}$
$=\frac{1}{70}$
Hence, the correct alternative is option (c).
View full question & answer→MCQ 501 Mark
A bag contain 4 white and 2 black balls. Two balls are drawn at random. The probability that they are of the same colour is ________.
- A
$\frac{5}{7}$
- B
$\frac{1}{7}$
- C
$\frac{7}{15}$
- D
$\frac{1}{15}$
Answer - $\frac{7}{15}$
Solution:
We assume that there are 4 white balls and 2 black balls.
There are $\big(\frac{6}{2}\big)=15$ total possible ways of drawing two balls from these given 6 balls.
We are interested in the event where the two drawn balls are of the same colour.
For this, we note that the number of ways of drawing 2 white balls is $\big(\frac{4}{2}\big)=6$ whereas the number of ways of drawing 2 black balls is$\big(\frac{2}{2}\big)=1.$
So, the probability that the two drawn balls are of the same colour is $\frac{6+1}{15}=\frac{7}{15}.$
View full question & answer→