Question

Let $p$ be the probability that a man aged $x$ years will die in a year time. The probability that out of ' $n$ ' men $A_1, A_2, A_3, \ldots . . ., A_n$ each aged ' $x$ ' years, $A_1$ will die \& will be the first to die is :

• A. $\frac{1-p^n}{n}$
• B. $\frac{ p }{ n }$
• C. $\frac{ p (1- p )^{ n -1}}{ n }$
• D. $\frac{1-(1-p)^n}{n}$

Answer 1

Answer: (D)

Probability that no body die in one year $=(1-p)^{ n }$
$\therefore \quad$ Probability that atleast one dies $=1-(1-P)^{ n }$. Since all the $n$ persons are equally likely to die.
Hence probability that $A_1$ is the 1 st person to $\operatorname{die}=\frac{1-(1-p)^n}{n}$
Alternatively
Probability that exactly one dies $={ }^n C_1 p(1-p)^{n-1}$
Since all the $n$ persons are equally likely to die
hence probability that $A_1 \operatorname{dies}=\frac{{ }^n C_1 p(1-p)^{n-1}}{n}$.......(1)
Probability that exactly two dies and $A _1$ is to die first $=\frac{{ }^{ n } C _2 p ^2(1- p )^{ n -2} \cdot{ }^{ n -1} C _1}{{ }^{ n } C _2} \cdot \frac{1}{2}$
$\left[\frac{{ }^{n-1} C_1}{{ }^n C_2}\right.$ is the probability of choosing $A_1$ and anyone out of $A_2$ to $A_n$ and out of these two, proabability that $A _1$ dies first $\left.=\frac{1}{2}\right]$
$=\frac{{ }^{ n } C _2 p ^2(1- p )^{ n -2}}{ n }$.....(2)
II ly Probability exactly 3 dies and $A _1$ is to die first $=\frac{{ }^{ n } C _3 p ^3(1- p )^{ n -3} \cdot{ }^{ a -1} C _2}{{ }^{ n } C _3} \cdot \frac{1}{3}$
$=\frac{{ }^n C_3 p^3(1-p)^3}{n}$ .....(3)
Hence Total probability (using $1- p = q$ )
$=\frac{{ }^n C _0 q ^{ n }+{ }^{ n } C _1 pq ^{ n -1}+{ }^{ n } C _2 p ^2 q ^{ n -2}+\ldots \ldots \ldots \ldots .+{ }^{ n } C _{ n } p ^{ n }-{ }^{ n } C _0 q ^{ n }}{ n } $
$=\frac{( q + p )^{ n }-(1- p )^{ n }}{ n }=\frac{1-(1- p )^{ n }}{ n }$