# Probability Questions

Q:

A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour.

 A) 1/2 B) 7/15 C) 8/15 D) 1/9

Explanation:

Let S be the sample space

Then n(S) = no of ways of drawing 2 balls out of (6+4) = $\inline&space;{\color{Black}10C_{2}}$ =$\inline \fn_cm \frac{10\times 9}{2\times 1}$ =45

Let E = event of getting both balls of same colour

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

=$\inline&space;{\color{Black}6C_{2}+4C_{2}}$ = $\inline \fn_cm \frac{6\times 5}{2\times 1}+\frac{4\times 3}{2\times 1}$= 15+6 = 21

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

74 14013
Q:

Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?

 A) 52/221 B) 55/190 C) 55/221 D) 19/221

Explanation:

We have n(s) = $\inline&space;{\color{Black}52C_{2}}$ = $\inline \fn_cm \frac{52\times 51}{2\times 1}$= 1326.

Let A = event of getting both black cards

B = event of getting both queens

A∩B = event of getting queen of black cards

n(A) = $\inline&space;{\color{Black}26C_{2}}$ = $\inline \fn_cm \frac{26\times 25}{2\times 1}$ = 325, n(B)= $\inline&space;{\color{Black}4C_{2}}$ = $\inline \fn_cm \frac{4\times 3}{2\times 1}$= 6  and  n(A∩B) = $\inline&space;{\color{Black}2C_{2}}$ = 1

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and

P(A∩B) = n(A∩B)/n(S) = 1/1326

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

33 8120
Q:

A man and his wife appear in an interview for two vacancies in the same post. The probability of husband's selection is (1/7) and the probability of wife's selection is (1/5). What is the probability that only one of them is selected ?

 A) 2/7 B) 1/7 C) 3/4 D) 4/5

Explanation:

64 7439
Q:

A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?

 A) 1/4 B) 1/2 C) 3/4 D) 7/12

Explanation:

Let A, B, C be the respective events of solving the problem and $\inline \overline{A},\overline{B},\overline{C}$ be the respective events of not solving the problem. Then A, B, C are independent events

$\inline \therefore \bar{A},\bar{B},\bar{C}$ are independent events

Now,  P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4

$\inline P(\bar{A})=\frac{1}{2},P(\bar{B})=\frac{2}{3} \: and\: P(\bar{C})=\frac{3}{4}$

$\inline \therefore$ P( none  solves the problem) = P(not A) and (not B) and (not C)

= $\inline P(\bar{A}\cap \bar{B}\cap \bar{C})$

= $\inline P(\bar{A})P(\bar{B})P(\bar{C})$             $\inline (\because \bar{A},\bar{B}\: and\: \bar{C}\: are\: independent)$                                  =  $\inline \frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}$

= $\inline \frac{1}{4}$

Hence, P(the problem will be solved) = 1 - P(none solves the problem)

= $\inline 1-\frac{1}{4}=\frac{3}{4}$

21 7072
Q:

In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

 A) 2/7 B) 5/7 C) 1/5 D) 1/2

$\inline \fn_jvn p(E)= \frac{n(E)}{n(S)}=\frac{10}{35}=\frac{2}{7}$