Question

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

Answer 1

Answer: Let's find the probability that at least 1 ball was red in both cases:

(A) When the first ball is replaced before the second draw:

Total number of balls = 4 red balls + 4 white balls = 8 balls.

Probability of drawing a red ball on the first draw = Number of red balls / Total number of balls = 4/8 = 1/2.

Since the ball is replaced before the second draw, the number of balls remains the same for the second draw.

Probability of drawing a red ball on the second draw (given that the first ball was red) = Number of red balls / Total number of balls = 4/8 = 1/2.

Now, to find the probability that at least 1 ball was red (either the first ball or the second ball or both), we need to find the complement of the probability that both balls were white:

Probability of drawing both balls white = (Number of white balls / Total number of balls) * (Number of white balls / Total number of balls) = (4/8) * (4/8) = 1/4.

Probability of at least 1 ball being red = 1 - Probability of drawing both balls white = 1 - 1/4 = 3/4.

So, the probability that at least 1 ball was red, given that the first ball was replaced before the second draw, is 3/4.

(B) When the first ball is not replaced before the second draw:

Total number of balls = 4 red balls + 4 white balls = 8 balls.

Probability of drawing a red ball on the first draw = Number of red balls / Total number of balls = 4/8 = 1/2.

Since the ball is not replaced before the second draw, there will be one less ball for the second draw.

Probability of drawing a red ball on the second draw (given that the first ball was red) = Number of red balls / (Total number of balls - 1) = 4/(8-1) = 4/7.

Now, to find the probability that at least 1 ball was red (either the first ball or the second ball or both), we again need to find the complement of the probability that both balls were white:

Probability of drawing both balls white = (Number of white balls / Total number of balls) * (Number of white balls - 1) / (Total number of balls - 1) = (4/8) * (3/7) = 3/14.

Probability of at least 1 ball being red = 1 - Probability of drawing both balls white = 1 - 3/14 = 11/14.

So, the probability that at least 1 ball was red, given that the first ball was not replaced before the second draw, is 11/14.