Probability Question 5
Question 5 - 2024 (29 Jan Shift 1)
A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws , is
(1) $\frac{5}{6}$
(2) $\frac{1}{6}$
(3) $\frac{5}{11}$
(4) $\frac{6}{11}$
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Answer (3)
Solution
Required probability $=$
$\frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^{3} \times \frac{1}{6}+\left(\frac{5}{6}\right)^{5} \times \frac{1}{6}+\ldots \ldots$
$=\frac{1}{6} \times \frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{11}$
