Answer:
(a)0.1317
(b)0.1646
(c)0.7901
Explanation:
This is a Binomial distribution problem.
For X ∼ Bin(n, p),
[TeX]P(X = k)=\left(\begin{array}{c}n\\ k \end{array} \right)p^{k}(1-p)^{n-k}[/TeX]
(a)All five people are still living.
The number of people still living has a Binomial(n = 5, p = 2/3)
[TeX]P(X = 5)=\left(\begin{array}{c}5\\ 5 \end{array} \right)(\frac{2}{3})^{5}(\frac{1}{3})^{0}[/TeX]
=[TeX](\frac{2}{3})^{5}[/TeX]
≈0.1317
(b)Exactly two people are still living
[TeX]P(X = 2)=\left(\begin{array}{c}5\\ 2 \end{array} \right)(\frac{2}{3})^{2}(\frac{1}{3})^{3}[/TeX]
[TeX]10 X (\frac{2}{3})^{2}(\frac{1}{3})^{3}[/TeX]
≈0.1646
(c) at least three people are still living.
P(X≥3)=P(X=3)+P(X=4)+P(X=5)
[TeX]P(X=3)=\left(\begin{array}{c}5\\ 3 \end{array} \right)(\frac{2}{3})^{3}(\frac{1}{3})^{2}[/TeX]
=0.3292
[TeX]P(X = 4)=\left(\begin{array}{c}5\\ 4 \end{array} \right)(\frac{2}{3})^{4}(\frac{1}{3})^{1}[/TeX]
=0.3292
[TeX]P(X = 5)=\left(\begin{array}{c}5\\ 5 \end{array} \right)(\frac{2}{3})^{5}(\frac{1}{3})^{0}[/TeX]
≈0.1317
P(X≥3)=0.3292+0.3292+0.1317
≈0.7901