Question

a) Determine the probability distribution for the number of women on a civil- court jury of 6, selected from a pool of 8 men and 10 women
b) what is the probability that there are at least 2 women in the jury ?

Answer 1

$$\binom{18}{6}={18! \over 6!(18-6)!}=18564$$

$$\binom{8}{0}={8! \over 0!(18-0)!}=1$$

$$\binom{8}{1}={8! \over 1!(18-1)!}=8$$

$$\binom{8}{2}={8! \over 2!(18-2)!}=28$$

$$\binom{8}{3}={8! \over 3!(18-3)!}=56$$

$$\binom{8}{4}={8! \over 4!(18-4)!}=70$$

$$\binom{8}{5}={8! \over 5!(18-5)!}=56$$

$$\binom{8}{6}={8! \over 6!(18-6)!}=28$$

$$\binom{10}{0}={10! \over 0!(10-0)!}=1$$

$$\binom{10}{1}={10! \over 1!(10-1)!}=10$$

$$\binom{10}{2}={10! \over 2!(10-2)!}=45$$

$$\binom{10}{3}={10! \over 3!(10-3)!}=120$$

$$\binom{10}{4}={10! \over 4!(10-4)!}=210$$

$$\binom{10}{5}={10! \over 5!(10-5)!}=252$$

$$\binom{10}{6}={10! \over 6!(10-6)!}=210$$

Let $X=$ the number of women on a civil- court jury of 6.

$$P(X=x)={\dbinom{10}{x}\dbinom{8}{6-x} \over \dbinom{18}{6}}$$

$$P(X=0)={\dbinom{10}{0}\dbinom{8}{6-0} \over \dbinom{18}{6}}={1(28) \over 18564}={1 \over 663}$$

$$P(X=1)={\dbinom{10}{1}\dbinom{8}{6-1} \over \dbinom{18}{6}}={10(56) \over 18564}={20 \over 663}$$

$$P(X=2)={\dbinom{10}{2}\dbinom{8}{6-2} \over \dbinom{18}{6}}={45(70) \over 18564}={75 \over 442}$$

$$P(X=3)={\dbinom{10}{3}\dbinom{8}{6-3} \over \dbinom{18}{6}}={120(56) \over 18564}={80 \over 221}$$

$$P(X=4)={\dbinom{10}{4}\dbinom{8}{6-4} \over \dbinom{18}{6}}={210(28) \over 18564}={70 \over 221}$$

$$P(X=5)={\dbinom{10}{5}\dbinom{8}{6-5} \over \dbinom{18}{6}}={252(8) \over 18564}={24 \over 221}$$

$$P(X=6)={\dbinom{10}{6}\dbinom{8}{6-6} \over \dbinom{18}{6}}={210(1) \over 18564}={5 \over 442}$$

Check

$${1 \over 663}+{20 \over 663}+{75 \over 442}+{80 \over 221}+{70 \over 221}+{24 \over221}+{5 \over 442}=1$$

The probability distribution for the number of women on a civil- court jury of 6

$$\def\arraystretch{1.5} \begin{array}{c: c: c: c: c: c:c: c: c: c} x_k & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline f(x_k) & {1 \over 663} & {20 \over 663} & {75 \over 442} & {80\over 221} & {70 \over 221} & {24 \over 221} & {5 \over 442} \end{array}$$

b) The probability that there are at least 2 women in the jury is

$$P(X\geq2)=1-P(X=0)-P(X=1)=$$

$$=1-{1 \over 663}-{20 \over 663}={214 \over 221}$$