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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:

Solution

✅ Correct Option: 1

Among 11 to 1919 there are 1010 odd and 99 even numbers.

The sum is even only when both are odd or both are even: (102)+(92)=45+36=81\binom{10}{2}+\binom{9}{2}=45+36=81.

Required conditional probability =4581=59=\frac{45}{81}=\frac{5}{9}.

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