Mathematics: 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

25.3 × is the equivalent expressionStep-by-step explanation:...

An insurance sales representative sells policies

Ответ:

quan91

11.01.2022

Mathematics

(a) $\frac{32}{243}$

(b) $\frac{192}{243}$

(d) $\frac{242}{243}$

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 $\frac{2}{3}$ . 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

$p = \frac{2}{3}$

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

q = 1 - p

$q = 1 - \frac{2}{3}$

$q = \frac{1}{3}$

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 × $(\frac{2}{3})^{5}$ × $(\frac{1}{3})^{5-5}$

= (1 × $(\frac{2}{3})^{5}$ × 1)

$=\frac{32}{243}$

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

p(x≥3) = ⁵C₃ × $(\frac{2}{3})^{3}$ × $(\frac{1}{3})^{5-3}$ + ⁵C₄ × $(\frac{2}{3})^{4}$ × $(\frac{1}{3})^{5-4}$ + ⁵C₅ × $(\frac{2}{3})^{5}$ × $(\frac{1}{3})^{5-5}$

p(x≥3) = (10 × $\frac{8}{27}$ × $\frac{1}{9}$ ) + (5 × $\frac{16}{81}$ × $\frac{1}{3}$ ) + (1 × $\frac{32}{243}$ × 1)

p(x≥3) $=\frac{80}{243}$ + $\frac{80}{243}$ + $\frac{32}{243}$ )

p(x≥3) $=\frac{192}{243}$

p(x=2) = ⁵C₂ × $(\frac{2}{3})^{2}$ × $(\frac{1}{3})^{5-2}$

p(x=2) = (10 × $\frac{4}{9}$ × $\frac{1}{27}$ )

p(x=2) = $\frac{40}{243}$

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

p(x≥1) = ⁵C₁ × $(\frac{2}{3})^{1}$ × $(\frac{1}{3})^{5-1}$ × ⁵C₂ × $(\frac{2}{3})^{2}$ × $(\frac{1}{3})^{5-2}$ × ⁵C₃ × $(\frac{2}{3})^{3}$ × $(\frac{1}{3})^{5-3}$ + ⁵C₄ × $(\frac{2}{3})^{4}$ × $(\frac{1}{3})^{5-4}$ + ⁵C₅ × $(\frac{2}{3})^{5}$ × $(\frac{1}{3})^{5-5}$

p(x≥1) = (5 × $\frac{2}{3}$ × $\frac{1}{81}$ ) × (10 × $\frac{4}{9}$ × $\frac{1}{27}$ ) × (10 × $\frac{8}{27}$ × $\frac{1}{9}$ ) + (5 × $\frac{16}{81}$ × $\frac{1}{3}$ ) + (1 × $\frac{32}{243}$ × 1)

p(x≥1) $=\frac{10}{243}$ + $\frac{40}{243}$ + $\frac{80}{243}$ + $\frac{80}{243}$ + $\frac{32}{243}$

p(x≥1) $=\frac{242}{243}$

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