Let's tackle each part of the question step by step.

Part (a):

What is the conditional probability of at least one die showing a 5, given that the sum is larger than 7?

This is a conditional probability problem where we are asked to find: $P(\text{at least one 5} \mid \text{sum} > 7)$ The formula for conditional probability is: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ Where:

• $A$ = "At least one die shows a 5"
• $B$ = "The sum is larger than 7"

First, we need to calculate $P(B)$, i.e., the probability that the sum of two dice is greater than 7, and $P(A \cap B)$, the probability that the sum is greater than 7 and at least one die shows a 5.

Total Sample Space:

The total sample space for rolling two dice is 36 (since each die has 6 faces, and $6 \times 6 = 36$).

Step 1: Calculate $P(B)$ (sum > 7)

The possible sums greater than 7 are:

• 8: (2,6), (3,5), (4,4), (5,3), (6,2)
• 9: (3,6), (4,5), (5,4), (6,3)
• 10: (4,6), (5,5), (6,4)
• 11: (5,6), (6,5)
• 12: (6,6)

Counting the outcomes, there are 15 outcomes where the sum is greater than 7. So: $P(B) = \frac{15}{36}$

Step 2: Calculate $P(A \cap B)$ (at least one 5 and sum > 7)

Now, we need to count the outcomes where at least one die is a 5, and the sum is greater than 7. These outcomes are:

• (3,5), (4,5), (5,3), (5,4), (5,5), (5,6), (6,5)

There are 7 such outcomes. So: $P(A \cap B) = \frac{7}{36}$

Step 3: Apply the conditional probability formula

Now, we can calculate $P(A \mid B)$: $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{7}{36}}{\frac{15}{36}} = \frac{7}{15}$

Part (a) Answer:

The conditional probability that at least one die shows a 5, given that the sum is greater than 7, is: $\boxed{\frac{7}{15}}$

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Part (b):

What is the conditional probability that the sum is larger than 7, given that at least one die is showing a 5?

We need to find: $P(\text{sum} > 7 \mid \text{at least one 5})$

Again, using the conditional probability formula: $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$ Where:

• $A$ = "At least one die shows a 5"
• $B$ = "Sum is larger than 7"

Step 1: Calculate $P(A)$ (at least one 5)

We need to count all outcomes where at least one die shows a 5. These outcomes are:

• (1,5), (2,5), (3,5), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,5)

There are 11 such outcomes. So: $P(A) = \frac{11}{36}$

Step 2: Recall $P(A \cap B)$ (at least one 5 and sum > 7)

We already calculated this in part (a), and it is: $P(A \cap B) = \frac{7}{36}$

Step 3: Apply the conditional probability formula

Now, we calculate $P(B \mid A)$: $P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{7}{36}}{\frac{11}{36}} = \frac{7}{11}$

Part (b) Answer:

The conditional probability that the sum is larger than 7, given that at least one die is showing a 5, is: $\boxed{\frac{7}{11}}$

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Do you want details or have any questions?

Here are some related questions to expand your understanding:

1. What is the probability of rolling a sum of exactly 7 with two dice?
2. What is the probability of getting doubles when rolling two dice?
3. What is the probability of getting at least one 6 when rolling two dice?
4. How would the problem change if one of the dice is weighted?
5. How do conditional probabilities apply to dependent and independent events?

Tip: Conditional probabilities are very useful when additional information helps refine the likelihood of an event occurring.