# There are 20 balls in the urn, and the number of white and black is the same. 4 balls were taken

**There are 20 balls in the urn, and the number of white and black is the same. 4 balls were taken out of the urn. What is the probability that there will be fewer than two black balls among the selected balls?**

There will be less than two black balls in the following cases: event A1 – took out all 4 white balls and event A2 – took out 1 black and 3 white balls. These are incompatible events and then event A, that among the taken out black balls less than two will be equal to the sum of these events.

A = A1 + A2;

Let’s find the probability of events A1 and A2 using the hypergeometric probability formula.

P (A1) = C (10.0) C (10.4) / C (20.4);

C (10.4) = 10! / (4! (10 – 4)!) = (10 9 8 7) / (1 2 3 4) = 210;

C (20.4) = 20! / (4! (20 – 4)!) = (20 19 18 17) / (1 2 3 4) = 4845;

C (10.0) = 1;

P (A1) = 1 210/4845 = 0.043;

P (A2) = C (10.1) C (10.3) / C (20.4);

C (10,1) = 10! / (1! (10 – 1)!) = 10;

C (10.3) = 10! / (3! (10 – 3)!) = (10 9 8) / (1 2 3) = 120;

P (A2) = 10 120/4845 = 0.248;

P (A) = P (A1) + P (A2) = 0.043 + 0.248 = 0.291.

Answer: The probability that among the selected black balls will be less than two is P (A) = 0.291.