There are 20 balls in the urn, 15 of them are white. 6 balls were taken out of the urn. What is the probability that there

There are 20 balls in the urn, 15 of them are white. 6 balls were taken out of the urn. What is the probability that there will be less than three white balls among the selected balls?

1. Let:

n = 20 balls in the urn;
n1 = 15 white balls;
n2 = 5 other balls;
k = 6 retrieved;
k1 <3 – white balls.
2. The number of outcomes favorable to the event X, which among the selected balls will be less than three white:

M = C (15, 1) * C (5, 5) + C (15, 2) * C (5, 4);
M = 15 * 1 + 105 * 5 = 15 + 525 = 540.
3. The total number of outcomes is equal to the number of combinations of all balls from 20 to 6:

N = C (20, 6);
N = 20! / (6! * 14!) = (20 * 19 * 18 * 17 * 16 * 15) / (1 * 2 * 3 * 4 * 5 * 6) = 19 * 17 * 8 * 15 = 323 * 120 = 38760.
4. Probability of event X:

P (X) = M / N = 540/38760 = 540 / (323 * 120) = 9 / (323 * 2) = 9/646 ≈ 0.0139.

Answer: 0.0139.