A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.

Solution

p = probability of success = $\frac{1}{6}$ , q = probability of failure = $\frac{5}{6}$

Third six comes at the 6 th throw so the remaining two sixes can appear in any of the previous 5 throws.

Probability of obtaining 2 sixes in 5 throws

$=^{5} C_{2} x \frac{1}{6} x \frac{1}{6} x \frac{125}{216}$

6 th throw definitely gives six with probability = $\frac{1}{6}$

Required probability

$= \frac{125}{216 x 36} x 10 x \frac{1}{6} = \frac{625}{23328}$

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