There are 11 white and 14 black balls in the urn. All balls are sequentially taken out of the urn.

There are 11 white and 14 black balls in the urn. All balls are sequentially taken out of the urn. Find the probability that 1) the white ball will be taken out third in order; 2) of the first three balls, at least one will be a white ball.

1) Options to choose from 3 balls, if the last one is white, and the probability of this event:

White, white, white – 11/25 * 10/24 * 9/23 = 990/13800.

White, black, white – 11/25 * 14/24 * 10/23 = 1540/13800.

Black, white, white – 14/25 * 11/24 * 10/23 = 1540/13800.

Black, black, white – 14/25 * 13/24 * 11/23 = 2002/13800.

Let’s sum up the probabilities: 990/13800 + 1540/13800 + 1540/13800 + 2002/13800 = 6072/13800 = 0.44.

Answer: the probability of choosing the white ball as the third in a row is 0.44.

2) Options for choosing balls, if at least 1 is white, and the probability of the event:

1 white and 2 black balls: C (1.11) * C (2.14) / C (3.25) = (11! / (1! * 10!) * 14! / (2! * 12!) ) / (25! / (3! * 22!)) = (11 * 13 * 14/2) / (23 * 24 * 25/6) = 6006/13800.

2 white and 1 black ball: C (2.11) * C (1.14) / C (3.25) = (11! / (2! * 9!) * 14! / (1! * 13!) ) / (25! / (3! * 22!)) = ((10 * 11/2) * 14) / (23 * 24 * 25/6) = 4620/13800.

3 white balls: C (3.11) / C (3.25) = (11! / (3! * 8!)) / (25! / (3! * 22!)) = (9 * 10 * 11 / 6) / (23 * 24 * 25/6) = 990/13800.

Now let’s calculate the sum of the probabilities: 6006/13800 + 4620/13800 + 990/13800 = 11616/13800 = 0.84.

Answer: the probability that out of 3 chosen balls, at least 1 will be white, is 0.84.