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A box contains 100 balls, numbered from 1 to 100

Expert replies
by Needgmat » Sat Oct 01, 2016 9:54 am
A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

A) 1/4

B) 3/8

C) 1/2

D) 5/8

E) 3/4

OAC

Please explain.
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Source: — Problem Solving |

by [email protected] » Sat Oct 01, 2016 10:03 am
Hi Needgmat,

These types of questions can be solved in a couple of different ways, depending on how you like to organize your information and how you "see" the question.

Since half the numbered balls are 'odd' and half are 'even', and we're replacing each ball after selecting it, on any given selection we have a 50/50 chance of getting an odd or even. This means that there are....

(2)(2)(2) = 8 possible arrangements when selecting 3 balls from the box:

EEE
EEO
EOE
OEE

OOO
OOE
OEO
EOO

Since we have the same number of Odds and Evens, each of these options has the same 1/8 probability of happening. Now you just have to find the ones that give us the ODD SUM that we're asked for. They are:

EEO
EOE
OEE
OOO

Four of the eight options. 4/8 = 1/2

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
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by ceilidh.erickson » Sat Oct 01, 2016 7:21 pm
Here's an even easier way to think about it: there are 100 terms between 1-100. Half of them are even, half are odd. As long as we have the same number of each, then any sum - whether it's of 3 numbers, or 4, or 5, etc - will have a 50/50 chance of being either even or odd.

To prove this, consider just the sum of 2 numbers:
OO
OE
EO
EE

Half of these (OO and EE) will have an even sum. The pattern will have to hold for the sum of any number of terms.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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