Question

A fair die is thrown 20 times. The probability that on the 10th throw, the fourth six appears is

Moderate

Solution

In the first 9 throws, we should have three sixes and six non-sixes; and a six in the 10th throw, and thereafter it does not matter whatever face appears. So, the required probability is
$^{9} C_{3} {(\frac{1}{6})}^{3} \times {(\frac{5}{6})}^{6} \times \frac{1}{6} \times \underset{10 times}{1 \times 1 \times 1 \times \dots \times 1} = \frac{84 \times 5^{6}}{6^{10}}$

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