# 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.

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