Two numbers are selected at random from the integers 1 through 19. If the sum is even, then the probability that both are odd numbers is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{5}{9} \)
From 1 to 19 there are 10 odd numbers and 9 even numbers. The sum of two numbers is even when both are odd or both are even.
\( \displaystyle \text{Both odd}:\ ^{10}C_{2}=45 \)
\( \displaystyle \text{Both even}:\ ^{9}C_{2}=36 \)
\( \displaystyle \text{Sum even}:\ 45+36=81 \)
This is a conditional probability, so the sample space shrinks to the 81 selections with an even sum:
\( \displaystyle P(\text{both odd}\mid\text{sum even})=\frac{45}{81} \)
\( \displaystyle =\frac{5}{9} \)