Question

Let $0.8$ is the probability that a man aged $90$ years will die in a year. If $p$ is the probability that out of $4$ men $A_1,A_2,A_3$ and $A_4$ each aged $90,A_1$ will die in a year and will be the first to die, then find $\frac{10000p}{48}$

Answer 1

Let $E_i$ denotes the event thst $A_i$ dies in a year.
Then $P\left(E_i\right)=0.8$ and $P\left(E_i^{\prime }\right)=0.2$ for $i=1,2,3,4$
Now, as all are independent event
Probability that none of them will die in a year $=0.2\times 0.2\times 0.2\times 0.2=(0.2{)}^4$
Let $E$ denotes the event that one of them will die in a year
$P\left(E\right)=1-$ (none of them will die in a year)
$P\left(E\right)=1-(0.2{)}^4=0.9984$
Let $F$ denotes the event that $A_1$ is the first one to die
$P\left(F\right)=\frac{1}{4}$
$\therefore$ Required probability that $A_1$ will die in a year and first one to die $p=\frac{1}{4}\times \left(0.9984\right)=0.2496$
$\therefore \frac{10000p}{48}=52$