One urn contains 5 white and 5 black balls. 8 balls are randomly removed from the urn.

One urn contains 5 white and 5 black balls. 8 balls are randomly removed from the urn. Find the probability that the 3 balls taken out of the urn are white.

1. Let:

n = 10 balls in the urn;
n1 = 5 white balls;
n2 = 5 black balls;
k = 8 taken out;
k1 = 3 of them are white;
k2 = 5 of them are black.
2. Let’s use the formula:

P = P (n1, k1, n2, k2) = С (n1, k1) * С (n2, k2) / C (n, k), where
C (n, k) = n! / (K! * (N – k)!) Are binomial coefficients.
3. The probability of occurrence of event A, consisting in the fact that 3 out of 8 taken out balls are white, is equal to:

P (A) = P (5, 3, 5, 5) = C (5, 3) * C (5, 5) / C (10, 8);
С (5, 3) = 5! / (3! * 2!) = 10;
C (5, 5) = 1;
С (10, 8) = 10! / (8! * 2!) = 45;
P (A) = 10 * 1/45 = 2/9.
Answer: 2/9.