Key Concepts and Formulas

• Conditional Probability: The probability of an event $F$ occurring given that another event $E$ has already occurred is denoted by $P(F|E)$.
• Formula for Conditional Probability: $P(F|E) = \frac{P(E \cap F)}{P(E)}$ Where $P(E \cap F)$ is the probability that both $E$ and $F$ occur, and $P(E)$ is the probability of event $E$.
• Simplified Formula for Equally Likely Outcomes: When all outcomes in the sample space are equally likely, the formula can be simplified by counting the number of favorable outcomes: $P(F|E) = \frac{n(E \cap F)}{n(E)}$ Here, $n(E)$ is the number of outcomes in event $E$, and $n(E \cap F)$ is the number of outcomes where both $E$ and $F$ occur. This method is often more direct for problems involving discrete events like dice rolls.

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Step-by-Step Solution

Step 1: Define the Sample Space and Events

• What we are doing: We first establish the complete set of all possible outcomes (the original sample space) and then clearly define the two specific events mentioned in the problem statement. This clarity is fundamental for correctly identifying and counting outcomes in subsequent steps.

• Why we are doing this: A precise definition of the sample space ensures we consider all possibilities, and a clear definition of events $E$ and $F$ prevents confusion about what we need to count for the numerator and denominator of our conditional probability formula.

• Original Sample Space ($S$): When a fair die is thrown two times, each throw can yield a score from 1 to 6. The total number of possible outcomes is $6 \times 6 = 36$. Each outcome is an ordered pair $(d_1, d_2)$, where $d_1$ is the score on the first die and $d_2$ is the score on the second die. For example, $(1,1), (1,2), \dots, (6,6)$.

• Event E (The Given Condition): "The sum of the scores appearing on the die is observed to be a multiple of 4." This is the event that has already occurred, and it will form our reduced sample space for the conditional probability.

• Event F (The Event of Interest): "The score 4 has appeared at least once." This is the event whose probability we need to find, but only within the context of Event E having occurred.

Our objective is to calculate $P(F|E)$, which, using the simplified formula, requires us to determine $n(E)$ and $n(E \cap F)$.

Step 2: Determine Outcomes for the Given Event (E)

• What we are doing: We systematically list all the pairs of scores $(d_1, d_2)$ from our original sample space whose sum is a multiple of 4. This set of outcomes constitutes Event E.
• Why we are doing this: The count of these outcomes, $n(E)$, will be the denominator in our conditional probability formula. It represents the size of our new, reduced sample space, as we are given that Event E has already happened.

Since $1 \le d_1 \le 6$ and $1 \le d_2 \le 6$:

• The minimum possible sum is $1+1=2$.
• The maximum possible sum is $6+6=12$.

Therefore, the possible multiples of 4 within the sum range [2, 12] are 4, 8, and 12. Let's list the outcomes for each possible sum:

• For Sum = 4:

  • $(1, 3)$
  • $(2, 2)$
  • $(3, 1)$ (3 outcomes)

• For Sum = 8:

  • $(2, 6)$
  • $(3, 5)$
  • $(4, 4)$
  • $(5, 3)$
  • $(6, 2)$ (5 outcomes)

• For Sum = 12:

  • $(6, 6)$ (1 outcome)

Combining these, the set of outcomes for Event E is: $E = \{ (1,3), (2,2), (3,1), (2,6), (3,5), (4,4), (5,3), (6,2), (6,6) \}$ The total number of outcomes in Event E is $n(E) = 3 + 5 + 1 = 9$.

Step 3: Determine Outcomes for the Intersection ($E \cap F$)

• What we are doing: We now identify which of the outcomes we found in Event E (where the sum is a multiple of 4) also satisfy Event F (where the score 4 has appeared at least once).
• Why we are doing this: The count of these outcomes, $n(E \cap F)$, will be the numerator in our conditional probability formula. It represents the number of outcomes that satisfy both the given condition and the event of interest.

Recall the outcomes of Event E: $E = \{ (1,3), (2,2), (3,1), (2,6), (3,5), (4,4), (5,3), (6,2), (6,6) \}$

Now, we check each outcome in set $E$ to see if it contains the score '4' at least once (i.e., satisfies Event F):

• $(1,3)$: Does not contain '4'.
• $(2,2)$: Does not contain '4'.
• $(3,1)$: Does not contain '4'.
• $(2,6)$: Does not contain '4'.
• $(3,5)$: Does not contain '4'.
• $\mathbf{(4,4)}$: Contains '4' (in both positions). This outcome satisfies Event F.
• $(5,3)$: Does not contain '4'.
• $(6,2)$: Does not contain '4'.
• $(6,6)$: Does not contain '4'.

The only outcome from set $E$ that also satisfies Event F is $(4,4)$. So, the intersection $E \cap F$ is: $E \cap F = \{ (4,4) \}$ The number of outcomes in $E \cap F$ is $n(E \cap F) = 1$.

Step 4: Calculate the Conditional Probability

• What we are doing: With $n(E)$ and $n(E \cap F)$ determined, we now substitute these values into the conditional probability formula.
• Why we are doing this: This is the final step to arrive at the solution to the problem, directly applying the formula derived from the key concepts.

Using the formula $P(F|E) = \frac{n(E \cap F)}{n(E)}$:

We found:

• $n(E) = 9$ (Number of outcomes where the sum is a multiple of 4)
• $n(E \cap F) = 1$ (Number of outcomes where the sum is a multiple of 4 AND the score 4 appeared at least once)

Substituting these values into the formula: $P(F|E) = \frac{1}{9}$

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Common Mistakes & Tips

• Confusing Events E and F: Always explicitly identify which event is the "given" condition (E, for the denominator) and which is the "event of interest" (F, for the intersection with E in the numerator).
• Incomplete Listing of Outcomes: For problems involving dice, systematically listing all possibilities for a given sum or condition helps avoid missing outcomes.
• Misinterpreting "At Least Once": The phrase "at least once" means one or more times. For a score of 4, it includes outcomes like $(4, x)$, $(x, 4)$, and $(4,4)$, where $x \ne 4$.
• Not Using the Reduced Sample Space: Remember that for conditional probability, the denominator is $n(E)$, not the total number of outcomes in the original sample space, $n(S)$.

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Summary

To find the conditional probability that the score 4 appeared at least once, given that the sum of the scores is a multiple of 4, we first identified all outcomes where the sum is a multiple of 4. This formed our reduced sample space, $E$, containing 9 outcomes. Next, from these 9 outcomes, we identified the ones where the score 4 appeared at least once, which was just the outcome $(4,4)$. This formed our intersection, $E \cap F$, containing 1 outcome. Finally, applying the conditional probability formula $P(F|E) = \frac{n(E \cap F)}{n(E)}$, we calculated the probability as $\frac{1}{9}$.

The final answer is $\boxed{\frac{1}{9}}$, which corresponds to option (B).