Respuesta :
Answer:
(a) [tex]\frac{32}{243}[/tex]
(b) [tex]\frac{192}{243}[/tex]
(c) [tex]\frac{40}{243}[/tex]
(d) [tex]\frac{242}{243}[/tex]
Additional Information:
An information is missing in the given question. The probability that a man of this particular age will be alive 30 years hence is [tex]\frac{2}{3}[/tex]. This additional information will be used to solve the question.
Step-by-step explanation:
Let the probability that one person will be alive in 30 years be p
[tex]p = \frac{2}{3}[/tex]
Let the probability that one person will not be alive in 30 years be q
q = 1 - p
[tex]q = 1 - \frac{2}{3}[/tex]
[tex]q = \frac{1}{3}[/tex]
The probability function is;
p(x) = ⁿCₓ × pˣ × q⁽ⁿ⁻ˣ⁾
(a) The probability that all 5 men will be alive in 30 years:
p(x=5) = ⁵C₅ × [tex](\frac{2}{3})^{5}[/tex] × [tex](\frac{1}{3})^{5-5}[/tex]
= (1 × [tex](\frac{2}{3})^{5}[/tex] × 1)
[tex]=\frac{32}{243}[/tex]
(b) The probability that at least 3 men will be alive in 30 years:
p(x≥3) = ⁵C₃ × [tex](\frac{2}{3})^{3}[/tex] × [tex](\frac{1}{3})^{5-3}[/tex] + ⁵C₄ × [tex](\frac{2}{3})^{4}[/tex] × [tex](\frac{1}{3})^{5-4}[/tex] + ⁵C₅ × [tex](\frac{2}{3})^{5}[/tex] × [tex](\frac{1}{3})^{5-5}[/tex]
p(x≥3) = (10 × [tex]\frac{8}{27}[/tex] × [tex]\frac{1}{9}[/tex]) + (5 × [tex]\frac{16}{81}[/tex] × [tex]\frac{1}{3}[/tex]) + (1 × [tex]\frac{32}{243}[/tex] × 1)
p(x≥3) [tex]=\frac{80}{243}[/tex] + [tex]\frac{80}{243}[/tex] + [tex]\frac{32}{243}[/tex])
p(x≥3) [tex]=\frac{192}{243}[/tex]
(c) The probability that only 2 men will be alive in 30 years:
p(x=2) = ⁵C₂ × [tex](\frac{2}{3})^{2}[/tex] × [tex](\frac{1}{3})^{5-2}[/tex]
p(x=2) = (10 × [tex]\frac{4}{9}[/tex] × [tex]\frac{1}{27}[/tex])
p(x=2) = [tex]\frac{40}{243}[/tex]
(d) The probability that at least 1 men will be alive in 30 years:
p(x≥1) = ⁵C₁ × [tex](\frac{2}{3})^{1}[/tex] × [tex](\frac{1}{3})^{5-1}[/tex] × ⁵C₂ × [tex](\frac{2}{3})^{2}[/tex] × [tex](\frac{1}{3})^{5-2}[/tex] × ⁵C₃ × [tex](\frac{2}{3})^{3}[/tex] × [tex](\frac{1}{3})^{5-3}[/tex] + ⁵C₄ × [tex](\frac{2}{3})^{4}[/tex] × [tex](\frac{1}{3})^{5-4}[/tex] + ⁵C₅ × [tex](\frac{2}{3})^{5}[/tex] × [tex](\frac{1}{3})^{5-5}[/tex]
p(x≥1) = (5 × [tex]\frac{2}{3}[/tex] × [tex]\frac{1}{81}[/tex]) × (10 × [tex]\frac{4}{9}[/tex] × [tex]\frac{1}{27}[/tex]) × (10 × [tex]\frac{8}{27}[/tex] × [tex]\frac{1}{9}[/tex]) + (5 × [tex]\frac{16}{81}[/tex] × [tex]\frac{1}{3}[/tex]) + (1 × [tex]\frac{32}{243}[/tex] × 1)
p(x≥1) [tex]=\frac{10}{243}[/tex] + [tex]\frac{40}{243}[/tex] + [tex]\frac{80}{243}[/tex] + [tex]\frac{80}{243}[/tex] + [tex]\frac{32}{243}[/tex]
p(x≥1) [tex]=\frac{242}{243}[/tex]
