A bag contains 5 white and 4 black balls. If 3 balls are drawn at random, what are theprobabilities of the following?(i) 2 of them are white? Ans : 10/21(ii) at most one of them is white Ans: 17/42(ii) at least two are white. Ans : 25/42

Sure, let's break down each of these probabilities:

(i) The probability that 2 of them are white:

First, we need to calculate the total number of ways to draw 3 balls from 9 (5 white and 4 black). This is a combination problem, so we use the combination formula: C(n, r) = n! / [r!(n-r)!], where n is the total number of items, r is the number of items to choose, and "!" denotes factorial.

So, the total number of ways to draw 3 balls from 9 is C(9, 3) = 9! / [3!(9-3)!] = 84 ways.

Next, we calculate the number of ways to draw 2 white balls from 5, and 1 black ball from 4. This is also a combination problem, so we use the combination formula again: C(5, 2) * C(4, 1) = 10 * 4 = 40 ways.

So, the probability that 2 of the balls are white is 40 / 84 = 10/21.

(ii) The probability that at most one of them is white:

This means we could have 0, 1, or 2 white balls. We calculate each of these probabilities separately and then add them together.

• The probability of 0 white balls: C(4, 3) / C(9, 3) = 4 / 84 = 1/21.
• The probability of 1 white ball: C(5, 1) * C(4, 2) / C(9, 3) = 60 / 84 = 10/14 = 5/7.
• The probability of 2 white balls: We've already calculated this as 10/21.

So, the total probability of at most one white ball is 1/21 + 5/7 + 10/21 = 17/42.

(iii) The probability that at least two are white:

This means we could have 2 or 3 white balls. We calculate each of these probabilities separately and then add them together.

• The probability of 2 white balls: We've already calculated this as 10/21.
• The probability of 3 white balls: C(5, 3) / C(9, 3) = 10 / 84 = 5/42.

So, the total probability of at least two white balls is 10/21 + 5/42 = 25/42.