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Out of $40$ consecutive integer $2$ are choosen at random , The the probability that their sum is odd, is

My Try::

Let We will take $40$ integer in that way.

$1,2,3,4,,...........................,40$

Now We choose $2$ out of $40$ is $\displaystyle = \binom{40}{2}$

Now We have to calculate probability for sum is even i.e $a+b = $Even.

Now we will break the $40$ consecutive integer into two parts.

$1,3,5,7,..........................,39$

$2,4,6,8,...........................40$

Now for sum is even we will take one from first row and one from second

Which can be done by $\displaystyle \binom{20}{1}.\binom{20}{1} = 20.20$

So Required probability is $\displaystyle = \frac{20}{39}$

Is this procedure is Right and answer given is $ = \displaystyle \frac{10}{39}$

Thanks

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3 Answers

up vote 0 down vote accepted

Your procedure is fine except for a "typo." You say that to get "sum is even" you pick one from first row and one from the second. What you mean is that to get "sum is odd" you pick one from first and one from second. If for example you pick $5$ and $18$, you get $23$, odd.

So the conclusion is that the probability the sum is odd is $\frac{20}{39}$.

It follows that the probability the sum is even is $1-\frac{20}{39}=\frac{19}{39}$.

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The last fraction is missing a 9 in the denominator. A probability go 6+ is quite big. –  Calvin Lin Jan 30 '13 at 18:47
    
Thanks! ${}{}{}{}{}{}$ –  André Nicolas Jan 30 '13 at 18:54
    
Thanks André Nicolas –  juantheron Jan 31 '13 at 3:52
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Although you can examine all $\frac{40 \cdot 39}2$ combinations, it is not necessary. Regardless of whether the first number chosen is even or odd, there are 19 of the remaining numbers which make an even sum and 20 which make an odd sum. Therefore you only need to compute the probabilities of the second number being in the appropriate category.

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Thanks half-integer fan –  juantheron Jan 31 '13 at 3:51
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No, this is not correct. You only get an even sum if you that two from the same row. You also change from wanting odd sum to wanting even sum in the middle, without saying you are going to subtract from $1$

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Thanks Ross Millikan –  juantheron Jan 31 '13 at 3:53
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