A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is
Solution
<p>Required probability $=$</p>
<p>$$\begin{aligned}
& \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^3 \times \frac{1}{6}+\left(\frac{5}{6}\right)^5 \times \frac{1}{6}+\ldots . . \\
& =\frac{1}{6} \times \frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{11}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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