A fair die is thrown until 2 appears. Then the probability that 2 appears in an even number of throws, is

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Solution

Q.
A fair die is thrown until 2 appears. Then the probability that 2 appears in an even number of throws, is [2024]

1 $\frac{1}{6}$

2 $\frac{6}{11}$

3 $\frac{5}{6}$

4 $\frac{5}{11}$

Ans.
(4)

Probability that 2 appears $= \frac{1}{6}$

Probability that 2 does not appear $= \frac{5}{6}$

Required probability $= \frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + \dots$

$= \frac{5}{6} \times \frac{1}{6} + {(\frac{5}{6})}^{3} \times \frac{1}{6} + {(\frac{5}{6})}^{5} \times \frac{1}{6} + \dots$ $= \frac{\frac{5}{6} \times \frac{1}{6}}{1 - \frac{5}{6} \times \frac{5}{6}}$ $= \frac{5}{11}$

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