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100 cubes were placed into a sack and each numbered from 1 [#permalink]

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17 Nov 2007, 16:40

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100 cubes were placed into a sack and each numbered from 1 to 100. If three cubes are to be withdrawn with replacement, What is the probability that the sum of the numbers on the withdrawn 3 cubes will be odd.

100 cubes were placed into a sack and each numbered from 1 to 100. If three cubes are to be withdrawn with replacement, What is the probability that the sum of the numbers on the withdrawn 3 cubes will be odd.

I get 1/4

two ways to get an odd number...

one odd and two even,
three odd

For each case, the chance is 1/8. Together, 2/8.

Or if we count the permutations - 1/2 - odd even even, odd odd odd, even odd even, even even odd.

Last edited by alrussell on 17 Nov 2007, 17:30, edited 1 time in total.

Why does the order of EVEN, ODD numbers matter when we are looking for sum of the numbers ?
Actually, with replacement makes me think that order does n't matter because the chances of EVEN and ODD numbers are equally likely...

100 cubes were placed into a sack and each numbered from 1 to 100. If three cubes are to be withdrawn with replacement, What is the probability that the sum of the numbers on the withdrawn 3 cubes will be odd.

OOO combination * probability +OEE combination * probability = 1*(50C1 *50C1*50C1/100C1*100C1*100C1) + 3!/2! *(50C1 *50C1*50C1/100C1*100C1*100C1) = 4 *1/8 =1/2
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100 cubes were placed into a sack and each numbered from 1 to 100. If three cubes are to be withdrawn with replacement, What is the probability that the sum of the numbers on the withdrawn 3 cubes will be odd.

Any integer has the same chance if replaced. so 1/2 should be answer. but with no replacement, it will be different.
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