A bag contains 4 red balls and 11 white balls. Two balls are drawn from the bag. What is the probability that the second ball is white W, given that the first ball is red and is replaced before the second is drawn?
A)Pr(
W) =
B)Pr(
W) =
C)Pr(
W) =
D)Pr(
W) =
E)Pr(
W) =

Question

A bag contains 4 red balls and 11 white balls. Two balls are drawn from the bag. What is the probability that the second ball is white W, given that the first ball is red and is replaced before the second is drawn? ​
A)Pr(
W) = 
B)Pr(
W) = 
C)Pr(
W) = 
D)Pr(
W) = 
E)Pr(
W) =

Quizplus · Accepted Answer

To find the probability of the second ball being white given that the first ball was red and is replaced before the second is drawn, we need to use the formula for conditional probability:
P(W|R) = P(W and R) / P(R)
where W and R denote the events of drawing a white and red ball respectively.

Since the first ball is replaced before the second is drawn, the probability of drawing a red ball on the first draw is 4 / 15 (since there are 4 red balls out of 15 total). 
The probability of drawing a white ball on the second draw, given that a red ball was drawn first and replaced, is 11 / 15 (since there are still 11 white balls out of 15 total).
The probability of both events happening (drawing a red ball first and a white ball second) is simply the product of the separate probabilities: (4 / 15) * (11 / 15) = 44 / 225.

Therefore, 
P(W|R) = (11 / 15) / (4 / 15) = 11 / 4 = 11eb0d3b_9827_5d00_9790_65e68dbf2971_TB1243_11.

A Bag Contains 4 Red Balls and 11 White Balls