Problem Statement
Two integers are selected at random from the first $20$ positive integers. If their sum is even, find the probability that both the numbers are odd.
Verified Solution & Marking Scheme
Define the Sample Space and Condition
First 20 positive integers: 10 odd numbers and 10 even numbers.
Two integers can be chosen in $\binom{20}{2} = \frac{20 \times 19}{2} = 190$ ways.
Let event $A$: Sum of two integers is even.
The sum is even if either both are odd OR both are even:
$\text{Number of ways} = \binom{10}{2} + \binom{10}{2} = 45 + 45 = 90$
Thus $P(A) = \frac{90}{190} = \frac{9}{19}$.
Define Event B and Intersection
Let event $B$: Both numbers are odd.
$B \cap A = \text{Both numbers are odd (since their sum is automatically even)}$
$\text{Number of favorable ways for } B \cap A = \binom{10}{2} = 45$
$P(B \cap A) = \frac{45}{190} = \frac{9}{38}$
Compute Conditional Probability P(B|A)
$P(B|A) = \frac{P(B \cap A)}{P(A)} = \frac{45/190}{90/190} = \frac{45}{90} = \frac{1}{2}$