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

The Correct Option is C

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: \(\frac{5}{11}\)