Question

An unbiased die is thrown again and again until three sixes are obtained. Find the probability of obtaining 3rd six in the sixth throw of the die.

Answer 1

Let p be the probability of obtaining a "six" in a single throw of the die. Then,

$p=\frac{1}{6}$ and $q=1-\frac{1}{6}=\frac{5}{6}$

Obtaining third six in the sixth throw of the die means that in first five throws there are 2 sixes and third six is obtained in sixth throw. Therefore,

Required probability = P( Getting 2 sixes in first 5 throws ) P( Getting 'six' in sixth throw )

$=({5_C}_2{p}^2{q}^5-2)\left(p\right)$

$={5_C}_2\left(\frac{1}{6}{)}^2\right(\frac{5}{6}{)}^3\times \frac{1}{6}$

$=\frac{625}{23328}$