Question

An urn contains 7 red, 5 white and 3 black balls. Three balls are drawn randomly one after the other without replacement. If it is known that first ball drawn is red and the second ball drawn is white, then the probability that the third ball drawn is not red is

• A. \( \frac{10}{13} \)
• B. \( \frac{8}{13} \)
• C. \( \frac{12}{13} \)
• D. \( \frac{7}{13} \)

Hint:

In conditional probability problems involving sequential draws without replacement, simply update the composition of the set (the urn, in this case) after each known event. The probability of the next event is then calculated based on this new composition.

Answer 1

The Correct Option is D

This is a conditional probability problem. We are given the outcomes of the first two draws.
Initially, the urn contains a total of \( 7+5+3 = 15 \) balls.
The first ball drawn is red. After this draw, the urn contains:
6 red, 5 white, and 3 black balls, for a total of 14 balls.
The second ball drawn is white. After this draw, the urn contains:
6 red, 4 white, and 3 black balls, for a total of 13 balls.
Now, we need to find the probability that the third ball drawn is *not* red, given the state of the urn.
The number of non-red balls remaining in the urn is the sum of white and black balls.
Number of non-red balls = 4 (white) + 3 (black) = 7.
The total number of balls remaining in the urn is 13.
The probability of drawing a non-red ball on the third draw is:
\( P(\text{3rd is not red}) = \frac{\text{Number of non-red balls}}{\text{Total number of balls}} = \frac{7}{13} \).