Question

Two digits are selected at random from the digits 1 through 9. If their sum is even, then the probability that both are odd is:

• A. \( \frac{3}{8} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{5}{8} \)
• D. \( \frac{3}{4} \)

Hint:

When calculating conditional probability, always focus on the specific condition given (in this case, the sum being even) and then find the favorable outcomes for the event of interest.

Answer 1

The Correct Option is C

We are selecting two digits from the numbers 1 through 9. Let’s break down the problem step by step. 
Step 1: Total number of ways to select two digits. The total number of ways to select two digits from the set \( \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \) is the number of combinations of 9 digits taken 2 at a time: \[ \binom{9}{2} = \frac{9 \times 8}{2} = 36. \] 
Step 2: Conditions for an even sum. For the sum of two digits to be even, either both digits must be even or both digits must be odd. Let's consider the number of ways these cases can happen. 
Case 1: Both digits are even. The even digits in the set are \( \{2, 4, 6, 8\} \), so the number of ways to select two even digits is: \[ \binom{4}{2} = \frac{4 \times 3}{2} = 6. \] 
Case 2: Both digits are odd. The odd digits in the set are \( \{1, 3, 5, 7, 9\} \), so the number of ways to select two odd digits is: \[ \binom{5}{2} = \frac{5 \times 4}{2} = 10. \] Thus, the total number of ways to select two digits such that their sum is even is: \[ 6 \text{ (both even)} + 10 \text{ (both odd)} = 16. \] 
Step 3: Probability that both digits are odd, given that their sum is even. We now need to find the probability that both digits are odd given that their sum is even. This is the conditional probability: \[ P(\text{both odd} \mid \text{sum even}) = \frac{\text{Number of ways to select both odd digits}}{\text{Total number of ways to select two digits with even sum}} = \frac{10}{16} = \frac{5}{8}. \] 
Thus, the correct answer is: \[ \boxed{\frac{5}{8}}. \]