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?


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.

  • $\begingroup$ would the answer be 1/3? $\endgroup$ – Daniel Mar 21 '12 at 16:57
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    $\begingroup$ The answer is not 1/3. It might be more useful if you would say what you tried; then someone might suggest where you went wrong. $\endgroup$ – MJD Mar 21 '12 at 16:59
  • $\begingroup$ is the naswer 5/9? $\endgroup$ – Daniel Mar 21 '12 at 17:07
  • $\begingroup$ I don't think this is an effective approach to the problem. $\endgroup$ – MJD Mar 21 '12 at 17:08
  • $\begingroup$ i took the probability of getting an odd number first then the probability of getting another odd from the remaining 8 numbers, then i did the same for even numbers. $\endgroup$ – Daniel Mar 21 '12 at 17:10

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 $\frac{4}{9}$.


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