Question

Two balls are drawn in succession out of a box containing 5 red and 4 white balls. Find the probability that at least 1 ball was​ red, given that the first ball was (Upper A )Replaced before the second draw. (Upper B )Not replaced before the second draw. ​(A) Find the probability that at least 1 ball was​ red, given that the first ball was replaced before the second draw. StartFraction 24 Over 49 EndFraction ​(Simplify your answer. Type an integer or a​ fraction.) ​(B) Find the probability that at least 1 ball was​ red, given that the first ball was not replaced before the second draw.

Answer 1

The answer is below

Explanation:

The box contains 5 red and 4 white balls.

A) The probability that at least 1 ball was​ red = P(both are red) + P(first is red and second is white) + P(first is white second is red)

Given that the first ball was (Upper A )Replaced before the second draw:

P(both are red) = P(red) × P(red) = 5/9 × 5/9 = 25/81

P(first is red and second is white) = P(red) × P(white) = 5/9 × 4/9 = 20/81

P(first is white and second is red) = P(white) × P(red) = 4/9 × 5/9 = 20/81

The probability that at least 1 ball was​ red = 25/81 + 20/81 + 20/81 = 65/81

B) The probability that at least 1 ball was​ red = P(both are red) + P(first is red and second is white) + P(first is white second is red)

Given that the first ball was not Replaced before the second draw:

P(both are red) = P(red) × P(red) = 5/9 × 4/8 = 20/72 (since it was not replaced after the first draw the number of red ball remaining would be 4 and the total ball remaining would be 8)

P(first is red second is white) = P(red) × P(white) = 5/9 × 4/8 = 20/72

P(first is white and second is red) = P(white) × P(red) = 4/9 × 5/8 = 20/72

The probability that at least 1 ball was​ red = 20/72 + 20/72 + 20/72 = 60/72