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?
Probability of an even sum
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probability
number-theory
2 Answers
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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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0@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 $\frac{4}{9}$.