Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both balls are red. (ii) first ball is black and second is red. (iii) one of them is black and other is red.
(a) P (both all red) \(=\frac{8}{18} \times \frac{8}{18}=\frac{16}{81}\)
(b) P (one of them is black and other red) = P(First ball black, second red) or P (First red, second black)
\(=\frac{10}{18} \times \frac{8}{18}+\frac{8}{18} \times \frac{10}{18}=\frac{20}{81}+\frac{20}{81}=\frac{40}{81}\)
Total number of balls = 18, number of red balls = 8 and number of black balls = 10
∴ Probability of drawing a red ball = 8/18
Similarly, probability of drawing a black ball = 10/18
(i) Probability of getting both red balls = P (both balls are red) = P (a red ball is drawn at first draw and again a red ball at second draw) = 8/18 x 8/18 = 16/81
(ii) P (probability of getting first ball is black and second is red) = 10/18 x 8/18 = 20/81
(iii) Probability of getting one black and other red ball = P(first ball is black and second is red) + P (first ball is red and second is black)
= 10/18 x 8/18 + 8/18 x 10/18
= 20/81 + 20/81 = 40/81