Question

In a sequence of independent trials, the probability of success is 1/4. If $p$ denotes the probability that the second success occurs on the fourth trial or later trial, find $32p$

Answer 1

Let the second success occur at the $n$th trial. This means that there was exactly one success in the first $n-1$ trials, so that the probability of getting the second success at the $n$th trial is
$p_n={(}^{n-1}{P}_{1}p{q}^{n-1-1})p=(n-1)p^2{q}^{n-2}$
Therefore the probability of the required event is
$P=p_4+p_5+p_6+.....$
$=p^2{q}^2(3+4q+5q^2+6q^3+....)$
$=p^2{q}^2\left[2\right(1+q+q^2+q^3+...)+q(1+2q+3q^2+....\left)\right]$
$=p^2{q}^2\left[3\right(1-q{)}^{-1}+q(1-q{)}^{-2}]1$
$=p^2{q}^2\left(\frac{3}{p}+\frac{q}{p^2}\right)=\frac{27}{32}$.
$\Rightarrow 32p=27$