Latest GRE Aptitude Question SOLUTION: two numbers are selected at random from the integers 1 through 9 the sum is even. what is the probability of both are odd?

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two numbers are selected at random from the integers 1 through 9 the sum is even. what is the probability of both are odd?

Asked In GRE siri (10 years ago)

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rocky balaji  (10 years ago)

1 2 3 4 5 6 7 8 9
chances that the sum is even= 41. since there are 81 possibilities(9*9) and 41 in them are even sum.
chances that both number are odd is 25(5*5)
therefore,probability= 25/41.

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venkatesh  (10 years ago)

the above submitted ans is wrong . this is correct

let two no's are a,b
so sum of a,b is even
sum of two no's is even only when two no's are either even or odd
even no : 2,4,6,8
possibilties are if a = 2 then b have 4 values 2,4,6,8 so 4 possobilities exists
so 4*4 = 16
odd no : 1,3,5,7,9
similarly
if a=1 then b has 5 values
so 5*5=25 posssibilities when both are odd
so prob(a,b are odd) = 25/(25+16)= 25/41

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Divya Tejaswi  (10 years ago)

ANS=5/8
Here it is just selection.So the order in which we take the numbers is not important.
Combinations of numbers which give even sum:
(1,3)(1,5)(1,7)(1,9)(3,5)(3,7)(3,9)(5,7)(5,9)(7,9)(2,4)(2,6)(2,8)(4,6)(4,8)(6,8)
There are totally 16 combinations
Out of these 10 pairs are odd numbers
Thus probability= 10/16 = 5/8
ANS= 5/8

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utkarsh singh  (10 years ago)

1 2 3 4 5 6 7 8 9
so odd no are 1 3 5 7 9
so no of cases with odd no =25
and total no of cases =81(9*9)
p=25/81

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siri  (10 years ago)

the answer is more than 3/8.

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venkatesh  (10 years ago)

let two no's are a,b
so sum of a,b is even
sum of two no's is even only when two no's are either even or odd
even no : 2,4,6,8
possibilties are if a = 2 then b have 3 values 4,6,8 so 3 possobilities exists
so 4*3 = 12
odd no : 1,3,5,7,9
similarly
if a=1 then b has 4 values
so 4*4=16 posssibilities when both are odd
so prob(a,b are odd) = 16/(16+12)= 4/7

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