# There are 5 white and 4 black balls in the urn. Two balls are taken out of the urn at random.

**There are 5 white and 4 black balls in the urn. Two balls are taken out of the urn at random. What is the probability that it will be a) two white balls b) two black balls c) one black and one white ball.**

1. The number of combinations of n1 white balls in k1 is equal to:

С (n1, k1).

2. The number of combinations of n2 black balls by k2 is equal to:

С (n2, k2).

3. The total number of combinations of n = n1 + n2 balls in k = k1 + k2:

C (n, k).

4. The probability of an event corresponding to the values of k1 and k2 is equal to the ratio of the number of favorable outcomes to the total number of outcomes:

P = P (n1, k1, n2, k2) = (С (n1, k1) * С (n2, k2)) / C (n, k).

For n1 = 5 and n2 = 4 we get:

P (5, k1, 4, k2) = (С (5, k1) * С (4, k2)) / C (9, k).

5. For the indicated events, we get:

a) two white balls;

k1 = 2; k2 = 0;

P1 = P (5, 2, 4, 0) = (C (5, 2) * C (4, 0)) / C (9, 2) = 10/36 = 5/18.

b) two black balls;

k1 = 0; k2 = 2;

P2 = P (5, 0, 4, 2) = (C (5, 0) * C (4, 2)) / C (9, 2) = 6/36 = 1/6.

c) one black and one white ball;

k1 = 1; k2 = 1;

P3 = P (5, 1, 4, 1) = (C (5, 1) * C (4, 1)) / C (9, 2) = 5 * 4/36 = 5/9.

Answer:

a) 5/18;

b) 1/6;

c) 5/9.