Question

Let 0.8 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, find 10000p

Answer 1

Let $E_i$ denote the event that $A_i$ dies in a year.
Then $P\left(E_i\right)=0.8$ and
$P\left(E_i\right)=0.2$ for $i=1,2,3,4$.
Now,P (none of $A_1,A_2,A_3,A_4$ dies in a year)
$=P(E_1^{\prime }\cap E_2^{\prime }\cap E_3^{\prime }\cap E_4^{\prime })=(0.2{)}^4$
Let E denote the event that at least one of $A_1,A_2,A_3,A_4$ dies in a year.
Then $P\left(E\right)=1-P(E_1^{\prime }\cap E_2^{\prime }\cap E_3^{\prime }\cap E_4^{\prime })=1-(0.2{)}^4=0.9984$.
If F denote the event that $A_1$ is first to die.
Then $P\left(F|E\right)=1/4$. Also,$p=\frac{1}{4}\left(0.9984\right)=0.2496$