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In a set of numbers there are 5 even numbers and 4 odd numbers. If two numbers are chosen at random from the set, without replacement, what is the probability that the sum of these two numbers is even?

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Hint: to get an even sum, you need two odds or two evens.

Added: to get two evens is $\frac 59 \cdot \frac 48$. Can you get the chance of two odds? As they are disjoint, you can add them.

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    @Daniel: Use Millikan hint, you will surely find your answer.2012-03-21
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I used the hint given by Ross Millikan answer.

P(the sum of the two numbers is even)=p(1st even and 2nd even)+p(1st odd and 2nd odd)

$\implies p(even sum)=\frac{5}{9}\times\frac{4}{8}+\frac{4}{9}\times \frac{3}{8}=\frac{20}{72}+\frac{12}{72}=\frac{4}{9} $

Therefore the probability that the sum of the two numbers is even is $\frac{4}{9}$.