Question

• A Box contains 3 blue balls and 5 yellow balls. A sample of 3 balls is selected from the Box without replacement. Let M be the number of blue balls contained in the sample, then find the probability distribution for M.

Answer 1

A sample can consist of 3 blue balls or of 2 blue and 1 yellow ball; of 1 blue and 2 yellow balls or of 3 yellow balls. The probability of first event is: $p_1=\frac{1}{C_8^3};$ $p_2=\frac{C_3^2C_5^1}{С_8^3}=\frac{C_3^2C_5^1}{С_8^3}$; $p_3=\frac{C_3^1C_5^2}{С_8^3}$; $p_4=\frac{C_5^3}{С_8^3}$, where $C_n^k=\frac{n!}{k!(n-k)!}.$