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
0
$\begingroup$
probability
number-theory
2 Answers
4
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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0would the answer be 1/3? – 2012-03-21
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1The 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. – 2012-03-21
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0is the naswer 5/9? – 2012-03-21
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0I don't think this is an effective approach to the problem. – 2012-03-21
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0i 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. – 2012-03-21
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0and i got 1/3 please help! – 2012-03-21
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0What probabilities did you take, and what did you do when you put them together, and what was the result? – 2012-03-21
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0first i took 5/9 then multiplied this with 1/2 – 2012-03-21
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0any help is appreaciated – 2012-03-21
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0@Daniel: Use Millikan hint, you will surely find your answer. – 2012-03-21
1
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 $\frac{4}{9}$.