Question

16 people are their in a college cricket team. 6 of them are batsman and 10 of them are bowlers. Two people are chosen at random. Calculate the probability that first is a batsman and second is bowler(without replacement).

• A. $\text{P(A and B)}=\frac{1}{4}$
• B. $\text{P(A and B)}=\frac{3}{8}$
• C. $\text{P(A and B)}=\frac{2}{3}$
• D. $\text{P(A and B)}=\frac{3}{4}$

Answer 1

The correct option is A $\text{P(A and B)}=\frac{1}{4}$

Given:
Number of people in the team $=16$

Number of batsman $=6$

Number of bowlers $=10$

Probability of choosing a batsman is $\text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

$\text{P(A)}=\frac{6}{16}=\frac{3}{8}$

Number of players left $=15$

Number of bowlers $=10$

Probability of choosing a bowler is $\text{P(B)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

$\text{P(B)}=\frac{10}{15}=\frac{2}{3}$

Probability of choosing two players in which one is batsman and other is bowler is $\text{P(A and B)}=\text{P(A)}\times \text{P(B)}$

$\text{P(A and B)}=\frac{3}{8}\times \frac{2}{3}=\frac{1}{4}$.