Question

A die is thrown again and again until three sixes are obtained. Find a if the probability of obtaining the third six in the sixth throw of the die is $\frac{a}{23328}$.

Answer 1

The probability of getting a six in a throw of die is $\frac{1}{6}$ and not getting a six is $\frac{5}{6}$.
Let $p=\frac{1}{6}$ and $q=\frac{5}{6}$
The probability that the 2 sixes come in the first five throws of the die is ${}^5{C}_2{\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^3=\frac{10\times {\left(5\right)}^3}{{\left(6\right)}^5}$
$\therefore$ Probability that third six comes in the sixth throw $=\frac{10\times {\left(5\right)}^3}{{\left(6\right)}^5}\times \frac{1}{6}$
$=\frac{10\times 125}{{\left(6\right)}^6}$
$=\frac{10\times 125}{46656}$

$=\frac{625}{23328}$

$=\frac{a}{23328}$

Hence, $a=625$.