Question

a jury pool has 24 men and 24 women, from which 12 jurors will be selected. assuming that each person is equally likely to be chosen and that the jury is selected at random, find the probability that the jury consists of
(a) all men
(b) all women
(c) 8 men and 4 women

Explanation:

Step1: Identify Total and Groups

Total people: \(24 + 24 = 48\). Probability of a man: \(p=\frac{24}{48}=0.5\), woman: \(q = 0.5\). We use binomial probability: \(P(X=k)=\binom{n}{k}p^{k}q^{n - k}\), \(n = 12\) (jurors).

Step2: Part (a) - All Men (\(k = 12\))

\(\binom{12}{12}(0.5)^{12}(0.5)^{0}=1\times(0.5)^{12}=\frac{1}{4096}\approx0.000244\)

Step3: Part (b) - All Women (\(k = 0\) men, \(k = 12\) women)

\(\binom{12}{0}(0.5)^{0}(0.5)^{12}=1\times1\times(0.5)^{12}=\frac{1}{4096}\approx0.000244\)

Step4: Part (c) - 8 Men, 4 Women (\(k = 8\))

\(\binom{12}{8}(0.5)^{8}(0.5)^{4}=\binom{12}{4}(0.5)^{12}\) (since \(\binom{n}{k}=\binom{n}{n - k}\)). \(\binom{12}{4}=\frac{12!}{4!8!}=495\). So \(495\times(0.5)^{12}=\frac{495}{4096}\approx0.1208\)

Answer:

s:
(a) \(\boldsymbol{\frac{1}{4096}\approx0.000244}\)
(b) \(\boldsymbol{\frac{1}{4096}\approx0.000244}\)
(c) \(\boldsymbol{\frac{495}{4096}\approx0.1208}\)