Step 1: Concept Explanation:
This problem involves conditional probability. We want to find the probability of event A (first throw is even) given that event B (sum of throws is 8) has occurred. This restricts our sample space to outcomes where the sum is 8.
Step 2: Formula:
The formula for conditional probability is:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\text{Number of outcomes in A and B}}{\text{Number of outcomes in B}} \]
Step 3: Detailed Solution:
Define the events:
Event B: Sum of the two throws is 8.
Event A: First throw is an even number.
First, list all pairs whose sum is 8, representing our reduced sample space (event B):
\[ B = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\} \]
The number of outcomes in B is \(n(B) = 5\).
Next, identify the outcomes in B where the first throw is even. These are the outcomes in \(A \cap B\).
From set B, the outcomes with an even number on the first throw are:
\[ A \cap B = \{(2, 6), (4, 4), (6, 2)\} \]
The number of outcomes in \(A \cap B\) is \(n(A \cap B) = 3\).
Now, calculate the conditional probability:
\[ P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{3}{5} \]
Step 4: Answer:
The probability of the first throw being even, given that the sum is 8, is \(\frac{3}{5}\).