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 determine the probability that the number 2 appears on an even-numbered throw of a fair die, we proceed as follows:

Key probabilities are:

• Probability of rolling a 2 on a single throw: \( \frac{1}{6} \).
• Probability of not rolling a 2 on a single throw: \( \frac{5}{6} \).

We define \(P(\text{Even})\) as the probability that the number 2 first appears on an even throw.

Consider the cases: if 2 appears on the 2nd throw, the first throw must not be 2. If 2 appears on the 4th throw, the first three throws must not be 2. This establishes a geometric progression of probabilities.

Specific probabilities:

• 2 appears on the 2nd throw: Event sequence is (not 2, 2). Probability is \( \left(\frac{5}{6}\right)^1 \times \frac{1}{6} = \frac{5}{36} \).
• 2 appears on the 4th throw: Event sequence is (not 2, not 2, not 2, 2). Probability is \( \left(\frac{5}{6}\right)^3 \times \frac{1}{6} = \frac{125}{1296} \).

The total probability \(P(\text{Even})\) is the sum of an infinite geometric 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 series has a first term \(a = \frac{5}{36}\) and a common ratio \(r = \left( \frac{5}{6} \right)^2\).

The sum of an infinite geometric series is calculated as:

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

Substituting the values:

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

Therefore, the probability is:

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

The probability that the number 2 appears on an even number of throws is \( \frac{5}{11} \).