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

To solve this problem, we need to determine the probability that the number 2 appears when a fair die is thrown an even number of times.

Consider the following: 

• The probability of rolling a 2 on a fair die in a single throw is \(\frac{1}{6}\).
• The probability of not rolling a 2 (i.e., rolling any of the other numbers) on a fair die is \(\frac{5}{6}\).

We're interested in the scenario where number 2 first appears on an even number of throws. Let's denote the probability that this happens as \(P(\text{Even})\).

To find \(P(\text{Even})\), consider that if 2 appears on the second throw, then the first throw must not be a 2. Similarly, if 2 appears on the fourth throw, then the first three throws must not be 2, and so on. These scenarios form a geometric progression.

Let's calculate the probability:

• For the 2nd throw: The event sequence is "not 2" then "2", with probability \(\left(\frac{5}{6}\right)^1 \times \frac{1}{6} = \frac{5}{36}\).
• For the 4th throw: The event sequence is "not 2", "not 2", "not 2", then "2", with probability \(\left(\frac{5}{6}\right)^3 \times \frac{1}{6} = \frac{125}{1296}\).

The probability \(P(\text{Even})\) can be expressed as the infinite series:

\[ P(\text{Even}) = \frac{5}{36} + \frac{125}{1296} + \cdots = \sum_{k=1}^{\infty} \left( \frac{5}{6} \right)^{2k-1} \times \frac{1}{6} \]

This is an infinite geometric series with the first term \(a = \frac{5}{36}\) and common ratio \(r = \left( \frac{5}{6} \right)^2\).

The sum of an infinite geometric series is given by:

\[ S = \frac{a}{1 - r} \]

Now substitute the values:

• \(a = \frac{5}{36}\)
• \(r = \left( \frac{5}{6} \right)^2 = \frac{25}{36}\)

Thus,

\[ P(\text{Even}) = \frac{\frac{5}{36}}{1 - \frac{25}{36}} = \frac{\frac{5}{36}}{\frac{11}{36}} = \frac{5}{11} \]

This calculation matches the given correct answer, so the final probability that 2 appears in an even number of throws is \(\frac{5}{11}\).

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: \(\frac{5}{11}\)