A committee of 5 members is to be formed from a group of 8 men and 5 women professionals. Let Y denote the number of women in the committee. If the selection is done randomly, determine the probability mass function of Y. What is the probability that the committee has at most 2 women?

Solution;

Let P(Y) be the probability of women in the committee;

$P(Y=0)=\frac{^5C_0 ×^8C_5}{^13C_5}=\frac{1×56}{1287}=\frac{56}{1287}$

$P(Y=1)=\frac{^5C_1×^8C_4}{^{13}C_5}=\frac{5×70}{1287}=\frac{350}{1287}$

$P(Y=2)=\frac{^5C_2×^8C_3}{^{13}C_5}=\frac{10×56}{1287}=\frac{560}{1287}$

$P(Y=3)=\frac{^5C_3×^8C_2}{^{13}C_5}=\frac{10×28}{1287}=\frac{280}{1287}$

$P(Y=4)=\frac{^5C_4×^8C_1}{^{13}C_5}=\frac{5×8}{1287}=\frac{40}{1287}$

$P(Y=5)=\frac{^5C_5×^8C_0}{^{13}C_5}=\frac{1×1}{1287}=\frac{1}{1287}$

The table for the probability mass function is;

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} Y & 0& 1&2&3&4&5\\ \hline P(Y=y) & \frac{56}{1287} & \frac{350}{1287}&\frac{560}{1287}&\frac{280}{1287} &\frac{40}{1287}&\frac{1}{1287}\\ \hdashline \end{array}$

The probability of at most 2 women is ;

$P(Y\leq2)=P(Y=0)+P(Y=1)+P(Y=2)$

$P(Y\leq2)=\frac{56}{1287}+\frac{350}{1287}+\frac{560}{1287}$

$P(Y\leq2)=\frac{966}{1287}$