JEE MAIN - Mathematics (2024 - 29th January Morning Shift - No. 6)
A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is
$$\frac{5}{11}$$
$$\frac{5}{6}$$
$$\frac{1}{6}$$
$$\frac{6}{11}$$
Explanation
Required probability $$=$$
$$\begin{aligned} & \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^3 \times \frac{1}{6}+\left(\frac{5}{6}\right)^5 \times \frac{1}{6}+\ldots . . \\ & =\frac{1}{6} \times \frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{11} \end{aligned}$$
Comments (0)
