Question

An insurance sales representative sells policies to 5 men, all of identical age and in good health. According to the actuarial tables, the probability that a man of this particular age will be alive 30 years hence is . Find the probability that in 30 years (a) all 5 men, (b) at least 3 men, (c) only 2 men, (d) at least 1 man will be alive

Answer 1

(a)
`(32)/(243)`

(b)
`(192)/(243)`

(c)
`(40)/(243)`

(d)
`(242)/(243)`

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
`(2)/(3)`. This additional information will be used to solve the question.

Explanation:

Let the probability that one person will be alive in 30 years be p

`p = (2)/(3)`

Let the probability that one person will not be alive in 30 years be q

q = 1 - p

`q = 1 - (2)/(3)`

`q = (1)/(3)`

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 ×
`((2)/(3))^(5)` ×
`((1)/(3))^(5-5)`

= (1 ×
`((2)/(3))^(5)` × 1)

`=(32)/(243)`

(b) The probability that at least 3 men will be alive in 30 years:

p(x≥3) = ⁵C₃ ×
`((2)/(3))^(3)` ×
`((1)/(3))^(5-3)` + ⁵C₄ ×
`((2)/(3))^(4)` ×
`((1)/(3))^(5-4)` + ⁵C₅ ×
`((2)/(3))^(5)` ×
`((1)/(3))^(5-5)`

p(x≥3) = (10 ×
`(8)/(27)` ×
`(1)/(9)`) + (5 ×
`(16)/(81)` ×
`(1)/(3)`) + (1 ×
`(32)/(243)` × 1)

p(x≥3)
`=(80)/(243)` +
`(80)/(243)` +
`(32)/(243)`)

p(x≥3)
`=(192)/(243)`

(c) The probability that only 2 men will be alive in 30 years:

p(x=2) = ⁵C₂ ×
`((2)/(3))^(2)` ×
`((1)/(3))^(5-2)`

p(x=2) = (10 ×
`(4)/(9)` ×
`(1)/(27)`)

p(x=2) =
`(40)/(243)`

(d) The probability that at least 1 men will be alive in 30 years:

p(x≥1) = ⁵C₁ ×
`((2)/(3))^(1)` ×
`((1)/(3))^(5-1)` × ⁵C₂ ×
`((2)/(3))^(2)` ×
`((1)/(3))^(5-2)` × ⁵C₃ ×
`((2)/(3))^(3)` ×
`((1)/(3))^(5-3)` + ⁵C₄ ×
`((2)/(3))^(4)` ×
`((1)/(3))^(5-4)` + ⁵C₅ ×
`((2)/(3))^(5)` ×
`((1)/(3))^(5-5)`

p(x≥1) = (5 ×
`(2)/(3)` ×
`(1)/(81)`) × (10 ×
`(4)/(9)` ×
`(1)/(27)`) × (10 ×
`(8)/(27)` ×
`(1)/(9)`) + (5 ×
`(16)/(81)` ×
`(1)/(3)`) + (1 ×
`(32)/(243)` × 1)

p(x≥1)
`=(10)/(243)` +
`(40)/(243)` +
`(80)/(243)` +
`(80)/(243)` +
`(32)/(243)`

p(x≥1)
`=(242)/(243)`