Question

Two fair dice, each with faces numbered$1,2,3,4,5$ and $6$, led 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

Find the required probability:

Step 1: Find the probability of given conditions

The sample size of the sum to be a prime number $=\left\{\right(1,1),(1,2),(2,1),(2,3),(3,2),(1,4),(4,1),(1,6),(2,5),(3,4),(4,3),(5,2),(6,1),(5,6),(6,5\left)\right\}=15$

The sample size of the sum to be an odd number is$=\left\{\right(1,2),(2,1),(2,3),(3,2),(1,4),(4,1),(1,6),(2,5),(3,4),(4,3),(5,2),(6,1),(5,6),(6,5\left)\right\}=14$

The sample size of the sum to be a perfect square$=\{(2,2),(3,1),(3,6),(1,3),(4,5),(5,4),(6,3)\}=7$

The probability of the sum is a prime number $=\frac{15}{36}$

The probability of the sum is a perfect square $=\frac{7}{36}$

Step 2: Find the value of $p$

The probability of the sum is a perfect square before being a prime number $=\frac{7}{36}+\left(\frac{15}{36}\right)\left(\frac{7}{36}\right)+{\left(\frac{15}{36}\right)}^2\left(\frac{7}{36}\right)+..........$

The probability of the sum is a perfect square which is an odd number ( Here, not mentioning the perfect square which is a perfect square before a prime number, talking about any perfect square)$=\frac{4}{36}+\left(\frac{14}{36}\right)\left(\frac{4}{36}\right)+{\left(\frac{14}{36}\right)}^2\left(\frac{4}{36}\right)+..........$

Now, the required probability $p$

$$\begin{gathered} =\frac{{\displaystyle \frac{4}{36}}+\left({\displaystyle \frac{14}{36}}\right)\left({\displaystyle \frac{4}{36}}\right)+{\left({\displaystyle \frac{14}{36}}\right)}^2\left({\displaystyle \frac{4}{36}}\right)+..........}{{\displaystyle \frac{7}{36}}+\left({\displaystyle \frac{15}{36}}\right)\left({\displaystyle \frac{7}{36}}\right)+{\left({\displaystyle \frac{15}{36}}\right)}^2\left({\displaystyle \frac{7}{36}}\right)+..........} \\ =\frac{4\left(\frac{1}{36}+\left(\frac{14}{36}\right)\left(\frac{1}{36}\right)+{\left(\frac{14}{36}\right)}^2\left(\frac{1}{36}\right)+.........\right)}{7\left(\frac{1}{36}+\left(\frac{15}{36}\right)\left(\frac{1}{36}\right)+{\left(\frac{15}{36}\right)}^2\left(\frac{1}{36}\right)+.........\right)} \\ =\frac{4}{7} \end{gathered}$$

So, the value of $14p$$=\frac{14\times 4}{7}=8$.

Hence, the correct answer is $8$.