Question

The probability that a man aged 60 will live to be 70 is 0.65. What is the probability that out of 10 men, now aged60 (i) exactly 9 will live to be 70 (ii) at most 9 will live to be 70, and (iii) at least 7 will live to be 70?

Answer 1

Answer on Question #45526, Engineering, SolidWorks — CosmoWorks — Ansys

The probability that a man aged 60 will live to be 70 is 0.65. What is the probability that out of 10 men, now aged 60 (i) exactly 9 will live to be 70 (ii) at most 9 will live to be 70, and (iii) at least 7 will live to be 70?

Solution

Here we are dealing with binomial distribution and hence we will use formula for probability mass function:

$f(k;n,p)=\Pr(X=k)={n\choose k}p^{k}(1-p)^{n-k}$

where n=60 is number of randomly selected men and p=0.65 is probability. (i)

exactly 9 will live to be 70. here k=9. so

$Pr(X=9)={10\choose 9}0.65^{9}(1-0.65)^{1}=0.0724917$

(ii)at most 9 will live to be 70. This can be find as 1 - Pr(only 1 survives to 70) - Pr(0 survived). Hence

$Pr(X\leq 9)=1-{10\choose 1}0.65^{1}(1-0.65)^{9}-{10\choose 0}0.65^{0}(1-0.65)^{10}=0.99946$

(iii) This can be find as

$Pr(X=7)+Pr(X=8)+Pr(X=9)+Pr(X=10)=0.513827$