Question

Two fair dice, each with faces numbered $1,2,3,4,5$ and $6,$ are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If $p$ is the probability that this perfect square is an odd number, then the value of $14p$ is

Answer 1

Sum is prime
$2:(1,1)$
$3:(1,2),(2,1)$
$5:(2,3),(3,2),(1,4),(4,1)$
$7:(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$
$11:(5,6),(6,5)$
$\therefore P\left(\text{prime}\right)=\frac{15}{36}$

Possible perfect squares are $4,9$
$4:(1,2),(2,1),(2,2)$
$9:(4,5),(5,4),(6,3),(3,6)$
$P\left(\text{perfect square}\right)=\frac{7}{36}$

Required probability,
$p=\frac{\frac{4}{36}+\left(\frac{14}{36}\right)\left(\frac{4}{36}\right)+{\left(\frac{14}{36}\right)}^2\left(\frac{4}{36}\right)+\cdots }{\frac{7}{36}+\left(\frac{14}{36}\right)\times \left(\frac{7}{36}\right)+{\left(\frac{14}{36}\right)}^2\left(\frac{7}{36}\right)+\cdots }$

$=\frac{4[\frac{1}{36}+\left(\frac{14}{36}\right)\left(\frac{1}{36}\right)+{\left(\frac{14}{36}\right)}^2\left(\frac{1}{36}\right)+\cdots ]}{7[\frac{1}{36}+\left(\frac{14}{36}\right)\times \left(\frac{1}{36}\right)+{\left(\frac{14}{36}\right)}^2\left(\frac{1}{36}\right)+\cdots ]}$

$\Rightarrow p=\frac{4}{7}$
$\therefore 14p=14\times \frac{4}{7}=8$