Question

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.

Answer 1

In conditional probability, If A and B are independent events, \(P(A\cap\;B)=P(A)\;P(B)\)
Given the total number of black balls = 10 and total number of red balls = 8.
P (drawing a black ball) = 10/18=5/9 and P (drawing a red ball) = 8/18=4/9
(i) P (both balls are red):
P (both balls are red) = P (a red is drawn first and a red is drawn second time also) = P (drawing a red) × P (drawing a red) = 4/9 × 4/9 = 16/81
(ii) P (first ball is black and second is red):
P (first ball is black and second is red) = P (drawing a black first) × P (drawing a red) = 5/9 × 4/9 = 20/81
(iii) P (getting one black and one red ball):
P (getting one black and one red ball) = P (first ball is black and second is red) + P (first ball is red and second is black)
We calculated the P (drawing a black ball first and red second above) in (ii). The P (drawing a red first and black second) is also the same.
Therefore, P (getting one black and one red ball) = 20/81+ 20/81=40/81