Conditional Probability of Rolling a 6 Exactly Twice

Introduction:
In this problem, we are given that a fair die is rolled three times independently. We need to find the conditional probability that the number 6 appeared exactly twice, given that it appeared at least once.

Approach:
We will use the concept of conditional probability to solve this problem. The conditional probability of an event A given another event B is denoted by P(A|B) and is calculated using the formula:

P(A|B) = P(A and B) / P(B)

To find the conditional probability that a 6 appeared exactly twice, given that it appeared at least once, we need to calculate the probability of two events:
1. Event A: A 6 appeared exactly twice.
2. Event B: A 6 appeared at least once.

Calculating the Probability of Event A:
To calculate the probability that a 6 appeared exactly twice, we need to consider all possible outcomes where exactly two 6's are rolled. Since we are rolling a fair die three times independently, there are a total of 6^3 = 216 possible outcomes.

Out of these 216 outcomes, we can choose two positions for the 6's in the sequence in 3 ways (e.g., 6-6-x, 6-x-6, x-6-6, where x is any number other than 6). For each of these positions, the remaining number can take any value from 1 to 5 (excluding 6) in 5 ways.

Therefore, the probability of event A is:

P(A) = (3 * 5) / 216 = 15 / 216

Calculating the Probability of Event B:
To calculate the probability that a 6 appeared at least once, we need to consider all possible outcomes where at least one 6 is rolled. We can calculate this probability as the complement of the probability that no 6 is rolled.

The probability of not rolling a 6 in one roll of the die is 5/6. Since we are rolling the die three times independently, the probability of not rolling a 6 in any of the three rolls is (5/6)^3 = 125/216.

Therefore, the probability of event B is:

P(B) = 1 - P(no 6 rolled) = 1 - (125/216) = 91/216

Calculating the Conditional Probability:
Now, we can calculate the conditional probability that a 6 appeared exactly twice given that it appeared at least once using the formula for conditional probability:

P(A|B) = P(A and B) / P(B)

We have already calculated P(A) as 15/216 and P(B) as 91/216.

To calculate P(A and B), we need to find the probability that a 6 appeared exactly twice and at least once. Since event A is a subset of event B, P(A and B) = P(A) = 15/216.

Therefore, the conditional probability is:

P(A|B) = (15/216) / (91/216) = 15/91

Conclusion:
The conditional probability that a 6 appeared exactly twice, given that it appeared at least once, is 15/91.