Two digits are selected at random from digits 1 through 9. If sum is even, find the probability that both numbers are odd?

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Two digits are selected at random from digits 1 through 9. If sum is even, find the probability that both numbers are odd?

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Out of the digits 1 through 9, 5 are odd and 4 are even.

The sum of the two digits is odd precisely when one digit is odd and the other digit is even, so

P(sum is odd) = P(odd, then even) + P(even, then odd).

It is not clear if these two digits are selected with or without replacement, so I will use both interpretations.

Without replacement, P(odd sum) = (5/9)(4/8) + (4/9)(5/8) = 5/9.

With replacement, P(odd sum) = (5/9)(4/9) + (4/9)(5/9) = 40/81.

Andha he kya ??ques thik se par

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