Key Concepts and Formulas

• Geometric Distribution: This distribution models the number of independent Bernoulli trials required to get the first success. If $p$ is the probability of success on a single trial and $q = 1-p$ is the probability of failure, then the probability that the first success occurs on the $k^{th}$ trial is given by: $P(X=k) = q^{k-1}p \quad \text{for } k=1, 2, 3, \ldots$
• Probability of Independent Events: If two events A and B are independent, the probability that both A and B occur is $P(A \text{ and } B) = P(A) \times P(B)$. This extends to multiple independent events.
• Sum of an Infinite Geometric Series: For an infinite geometric series $a + ar + ar^2 + \ldots$, its sum $S$ is given by: $S = \frac{a}{1-r}$ This formula is valid if and only if the absolute value of the common ratio, $|r|$, is less than 1 ($|r| < 1$). Here, $a$ is the first term and $r$ is the common ratio.

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Step-by-Step Solution

Step 1: Define Probabilities of Success and Failure for a Single Throw

First, we identify the probabilities associated with a single throw of a fair die.

• A fair die has 6 equally likely outcomes: $\{1, 2, 3, 4, 5, 6\}$. Each outcome has a probability of $\frac{1}{6}$.
• Success (S): Getting a '2'.
  • The probability of success in a single throw is $p = P(\text{getting a 2}) = \frac{1}{6}$.
• Failure (F): Not getting a '2'.
  • The probability of failure in a single throw is $q = P(\text{not getting a 2}) = 1 - p$.
  • $q = 1 - \frac{1}{6} = \frac{5}{6}$.
  • Alternatively, the outcomes for failure are $\{1, 3, 4, 5, 6\}$, which are 5 out of 6 outcomes, so $q = \frac{5}{6}$.

Reasoning: These probabilities ($p$ and $q$) are fundamental because each die throw is an independent Bernoulli trial. They will be used to calculate the probability of sequences of successes and failures.

Step 2: Identify the Events Leading to '2' Appearing in an Even Number of Throws

We are interested in the probability that the first '2' appears on an even-numbered throw. This means the first '2' can appear on the 2nd throw, or the 4th throw, or the 6th throw, and so on. These are mutually exclusive events, so their probabilities can be summed. Let $X$ be the random variable representing the number of throws until the first '2' appears. We want to find $P(X \text{ is even})$.

Using the Geometric Distribution formula $P(X=k) = q^{k-1}p$:

• Case 1: The first '2' appears on the 2nd throw ($X=2$).

  • This implies the 1st throw is a failure ($q$) and the 2nd throw is a success ($p$).
  • $P(X=2) = q^{2-1}p = qp = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$

• Case 2: The first '2' appears on the 4th throw ($X=4$).

  • This implies the 1st, 2nd, and 3rd throws are failures ($q^3$) and the 4th throw is a success ($p$).
  • $P(X=4) = q^{4-1}p = q^3p = \left(\frac{5}{6}\right)^3 \times \frac{1}{6} = \frac{125}{216} \times \frac{1}{6} = \frac{125}{1296}$

• Case 3: The first '2' appears on the 6th throw ($X=6$).

  • This implies the first five throws are failures ($q^5$) and the 6th throw is a success ($p$).
  • $P(X=6) = q^{6-1}p = q^5p = \left(\frac{5}{6}\right)^5 \times \frac{1}{6} = \frac{3125}{7776} \times \frac{1}{6} = \frac{3125}{46656}$

Reasoning: We are applying the definition of Geometric Distribution to find the probability of the first success occurring at specific even-numbered trials. Each term represents a sequence of failures followed by a single success.

Step 3: Formulate the Infinite Geometric Series

The total probability that '2' appears in an even number of throws is the sum of the probabilities of these mutually exclusive events: $P(\text{even throws}) = P(X=2) + P(X=4) + P(X=6) + \ldots$ Substituting the general expressions from Step 2: $P(\text{even throws}) = (qp) + (q^3p) + (q^5p) + \ldots$ Now, we substitute the specific values of $p = \frac{1}{6}$ and $q = \frac{5}{6}$: $P(\text{even throws}) = \left(\frac{5}{6} \times \frac{1}{6}\right) + \left(\left(\frac{5}{6}\right)^3 \times \frac{1}{6}\right) + \left(\left(\frac{5}{6}\right)^5 \times \frac{1}{6}\right) + \ldots$ This is an infinite geometric series. To find its sum, we need to identify its first term ($a$) and its common ratio ($r$).

• First Term ($a$): The first term of this specific series is $P(X=2)$. $a = qp = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$

• Common Ratio ($r$): The ratio of any term to its preceding term. Let's take the ratio of the second term to the first term: $r = \frac{q^3p}{qp} = q^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$ (We can verify this by checking the ratio of the third term to the second term: $\frac{q^5p}{q^3p} = q^2 = \frac{25}{36}$.) The common ratio is $r = \frac{25}{36}$.

Reasoning: We are expressing the desired total probability as a sum of specific probabilities, recognizing this sum as an infinite geometric series. Correctly identifying the first term ($a$) and common ratio ($r$) for this specific series is critical for applying the sum formula.

Step 4: Calculate the Sum of the Infinite Geometric Series

The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$, provided that $|r| < 1$. In our case, $a = \frac{5}{36}$ and $r = \frac{25}{36}$. Since $|r| = \left|\frac{25}{36}\right| < 1$, the sum converges, and we can use the formula.

Substitute the values of $a$ and $r$ into the formula: $P(\text{even throws}) = S = \frac{a}{1 - r} = \frac{\frac{5}{36}}{1 - \frac{25}{36}}$ First, simplify the denominator: $1 - \frac{25}{36} = \frac{36}{36} - \frac{25}{36} = \frac{36 - 25}{36} = \frac{11}{36}$ Now, substitute this back into the sum formula: $S = \frac{\frac{5}{36}}{\frac{11}{36}}$ $S = \frac{5}{36} \times \frac{36}{11}$ $S = \frac{5}{11}$

Reasoning: This step completes the calculation by applying the standard formula for the sum of an infinite geometric series, which efficiently sums all the infinite possibilities for the first '2' appearing on an even throw.

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Common Mistakes & Tips

• Incorrect Common Ratio: A common error is to mistake the common ratio of the series for the probability of failure $q$. In this problem, because we are summing probabilities for $X=2, 4, 6, \ldots$, the exponents of $q$ increase by 2, making the common ratio $q^2$, not $q$.
• Misidentifying the First Term: Ensure the first term 'a' of the series is correctly identified as $P(X=2)$, not $P(X=1)$ or just $p$.
• Understanding Geometric Distribution: Remember that $P(X=k) = q^{k-1}p$ means there are $k-1$ failures before the first success on the $k^{th}$ trial.

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Summary

This problem is a classic application of the Geometric Distribution combined with the summation of an infinite geometric series. By first identifying the probability of success ($p$) and failure ($q$) for a single die throw, we then determined the probabilities for the first '2' appearing on the 2nd, 4th, 6th, and subsequent even-numbered throws. These probabilities form an infinite geometric series. The key to solving the problem was correctly extracting the first term ($a = qp$) and the common ratio ($r = q^2$) of this series and then applying the sum formula $S = \frac{a}{1-r}$.

The final answer is $\boxed{\frac{5}{11}}$, which corresponds to option (A).