There are 10 balls in the urn, of which 2 are white, 3 are black and 5 are blue. 3 balls are drawn at random.

There are 10 balls in the urn, of which 2 are white, 3 are black and 5 are blue. 3 balls are drawn at random. What is the probability that all 3 balls are different colors?

1. Let:

n = 10 balls in the urn;
n1 = 2 white balls;
n2 = 3 black balls;
n3 = 5 blue balls;
k = 3 retrieved;
k1 = 1 white ball;
k2 = 1 black ball;
k3 = 1 blue ball.
2. The number of outcomes favorable to event X, that all 3 balls are of different colors:

a) the number of combinations of white balls from n1 to k1:

M1 = C (n1, k1) = C (2, 1) = 2;

b) the number of combinations of black balls from n2 to k2:

M2 = C (n2, k2) = C (3, 1) = 3;

c) the number of combinations of blue balls from n3 to k3:

M3 = C (n3, k3) = C (5, 1) = 5;

M = M1 * M2 * M3 = 2 * 3 * 5 = 30.

3. Total number of outcomes – the number of combinations of all balls from 10 to 3:

N = C (n, k) = C (10, 3) = 10! / (3! * 7!) = (10 * 9 * 8) / (1 * 2 * 3) = 10 * 3 * 4 = 120 …

4. Probability of event X:

P (X) = M / N = 30/120 = 1/4.

Answer: 1/4.