Question

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

• A. $\frac{5}{6}$
• B. $\frac{1}{6}$
• C. $\frac{5}{11}$
• D. $\frac{6}{11}$

Answer 1

Answer: (C)

$\frac{5}{11}$

Answer 2

Define the Probability of Success and Failure:
- The probability of rolling a 2 on any single throw is $\frac{1}{6}$.
- The probability of not rolling a 2 is $\frac{5}{6}$.

Calculate the Required Probability:
For a 2 to appear in an even number of throws, we consider the probabilities that it first appears on the 2nd, 4th, 6th, etc., throw. The probability of 2 appearing on the $2n$-th throw (even throws) is:
$\left( \frac{5}{6} \right)^{2n-1} \times \frac{1}{6}$
The required probability is an infinite series:
$\frac{5}{6} \times \frac{1}{6} + \left( \frac{5}{6} \right)^3 \times \frac{1}{6} + \left( \frac{5}{6} \right)^5 \times \frac{1}{6} + \dots$

Summing the Series:
This is a geometric series with the first term $\frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$ and common ratio $\left( \frac{5}{6} \right)^2 = \frac{25}{36}$. The sum of an infinite geometric series is given by:
$\text{Sum} = \frac{\text{first term}}{1 - \text{common ratio}} = \frac{\frac{5}{36}}{1 - \frac{25}{36}} = \frac{\frac{5}{36}}{\frac{11}{36}} = \frac{5}{11}$
So, the correct option is: $\frac{5}{11}$