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by danjuma » Sat Sep 25, 2010 11:40 am
A box contains 100 balls numbered from 1 to 100. If 3 balls are selected at random and with replacement from the box, what is the probability that the sum of the 3 numbers on the balls selected will be odd.

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by vijaynaik » Sat Sep 25, 2010 1:27 pm
[spoiler]1/4[/spoiler]

Probability of picking an ODD#or an EVEN# is 50/100 = 1/2.

We need to pick 3 ODD#s OR 1 ODD and 2 EVEN#s to get ODD sum.

case1. P(3 ODD#s)= 1/2 *1/2* 1/2 = 1/8
case2. P(1 odd and 1 even and 1 even) = 1/2 * 1/2 * 1/2 = 1/8

P(ODD sum) = [spoiler]1/8 + 1/8 = 1/4;[/spoiler]
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by GMATGuruNY » Sat Sep 25, 2010 2:34 pm
danjuma wrote:A box contains 100 balls numbered from 1 to 100. If 3 balls are selected at random and with replacement from the box, what is the probability that the sum of the 3 numbers on the balls selected will be odd.

Please note-no answer choices available.

Thanks.
The sum will be odd if all 3 numbers selected are odd or if exactly 1 number selected is odd.
Let O = odd and E = even.
Since there are an equal number of odd numbers and even numbers in the box, P(O) = 1/2 and P(E) = 1/2.

Probability all 3 are odd:
P(OOO) = 1/2 * 1/2 * 1/2 = 1/8

Probability exactly 1 is odd:
P(OEE) = 1/2 * 1/2 * 1/2 = 1/8
Since EOE and EEO are also good outcomes, we multiply this result by 3:
3 * 1/8 = 3/8

Since we'll get a good outcome if all 3 numbers selected are odd OR if exactly 1 number selected is odd, we add the fractions:
1/8 + 3/8 = 4/8 = 1/2.
Last edited by GMATGuruNY on Sat Sep 25, 2010 5:25 pm, edited 1 time in total.
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by vijaynaik » Sat Sep 25, 2010 2:49 pm
Thanks GMATGuruNY.
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