Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

Solution:

Total number of balls = 18

Number of red balls = 8

Number of black balls = 10

(i) Probability of getting a red ball in the first draw $=\frac{8}{18}=\frac{4}{9}$

The ball is replaced after the first draw.

$\therefore$ Probability of getting a red ball in the second draw $=\frac{8}{18}=\frac{4}{9}$

Therefore, probability of getting both the balls red $=\frac{4}{9} \times \frac{4}{9}=\frac{16}{81}$

(ii) Probability of getting first ball black $=\frac{10}{18}=\frac{5}{9}$

The ball is replaced after the first draw.

Probability of getting second ball as red $=\frac{8}{18}=\frac{4}{9}$

Therefore, probability of getting first ball as black and second ball as red $=\frac{5}{9} \times \frac{4}{9}=\frac{20}{81}$

(iii) 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 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

$=\frac{20}{81}+\frac{20}{81}$

$=\frac{40}{81}$