gauravsoni wrote:

A barrel contains only red balls, white balls, and brown balls. If two balls are selected at random without replacing each ball, what is the probability that two brown balls will be selected?

(1) The probability that one ball selected at random is either red or white is .

(2) There are nine balls in the barrel.

This is an interesting one. The question says we have only red, white and brown balls. And we have to find the probability that both balls drawn with replacement are brown.

Probability of getting 1 brown ball = number of brown balls/ total number of balls = A

Probability that both balls drawn are brown = A.A = A^2. (Note that A is dependent on number of brown balls and total number of balls which are 2 variables)

Statement 1: The probability that one ball selected at random is either red or white is 2/3.

This means probability of not getting a brown ball =2/3

Therefore, A which is probability of getting a brown ball is 1/3

Now since, we want to get A^2, it may seem that we can answer the question. But this is a trap. Lets say, we have only 3 balls. A=1/3 means, we have only 1 brown ball. Thus, probability of getting 2 brown balls become 0. In case it is more than 3 balls, i.e. 6, 9, 12, ans so on, we can predict the probability.

Hence, this is insufficient

Statement 2: There are nine balls in the barrel.

We don't know the number of brown balls and hence can't predict the probability.

Hence insufficient.

Now, let us take both the statement together. After statement 1, if we can assure that the barrel has more than 3 balls, we can definitely tell the probability. Now, we are given 9 balls. Which means we can calculate the probability.

Hence, answer is C.

Hope it helps!!

Kudos if you like!!