Assume the box contains 7 balls:
3 red, 2 blue, and 2 yellow. A ball is drawn and its
color noted. If the ball is yellow, it is replaced; otherwise, it is not.
A second ball is then drawn and its color is noted.
What is the probability that the first ball was yellow, given that the
second was red?
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To find the probability that the first ball was yellow, given that the second ball was red, we need to use conditional probability.
Let's break down the steps to find the solution:
1. Determine the total number of possible outcomes:
The first ball could be any of the 7 balls in the box, and the second ball could also be any of the remaining 6 balls. So the total number of possible outcomes is 7 * 6 = 42.
2. Determine the number of favorable outcomes:
We want to find the situations where the first ball drawn is yellow and the second ball drawn is red. There are two cases to consider:
a) Case 1: The first ball drawn is yellow and the second ball drawn is red.
b) Case 2: The first ball drawn is not yellow (red or blue) and the second ball drawn is red.
Let's calculate each case separately:
a) For Case 1, the probability of drawing a yellow ball and then a red ball is (2/7) * (3/6) = 1/7.
b) For Case 2, the probability of drawing a non-yellow ball and then a red ball is (5/7) * (3/6) = 5/14.
So the number of favorable outcomes for this specific scenario is 1/7 + 5/14 = 3/7.
3. Calculate the conditional probability:
The conditional probability is given by the formula:
P(A|B) = P(A intersection B) / P(B)
In this case, A represents the event "the first ball drawn is yellow" and B represents the event "the second ball drawn is red."
P(A intersection B) is the probability of both events happening together, which is 1/7 (found in step 2).
P(B) is the probability of the second ball drawn being red. Since we are not replacing the ball when it is yellow, the total number of possible outcomes for the second ball changes to 5. So P(B) = 3/5.
Therefore, the conditional probability, P(A|B), is:
P(A|B) = (1/7) / (3/5) = 5/21.
So, the probability that the first ball was yellow, given that the second was red, is 5/21.