Question

Two numbers are selected at random from integers 1 to 9. If their sum is even, what is the probability that both the numbers are odd?

• A. \( \frac{4}{9} \)
• B. \( \frac{5}{8} \)
• C. \( \frac{1}{6} \)
• D. \( \frac{2}{3} \)

Hint:

Even sum occurs when both numbers are of same parity (both even or both odd).

Answer 1

The Correct Option is B

Step 1: Count total numbers.
Integers from 1 to 9:
Odd numbers = \( \{1,3,5,7,9\} \) → 5 numbers
Even numbers = \( \{2,4,6,8\} \) → 4 numbers

Step 2: Condition for sum to be even.
Sum is even when:
- Both numbers are even OR both are odd

Step 3: Count favorable cases (sum even).
Number of ways to choose 2 odd numbers:
\[ {}^5C_2 = 10 \]
Number of ways to choose 2 even numbers:
\[ {}^4C_2 = 6 \]
Total favorable cases (sum even):
\[ 10 + 6 = 16 \]

Step 4: Required cases (both odd).
\[ {}^5C_2 = 10 \]

Step 5: Conditional probability.
\[ P(\text{both odd} \mid \text{sum even}) = \frac{10}{16} \]

Step 6: Simplify.
\[ = \frac{5}{8} \]

Step 7: Final Answer.
\[ \boxed{\frac{5}{8}} \]