Practice question for mba.com preparatory material

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pranavc: Master | Next Rank: 500 Posts; Posts: 167; Joined: Tue Feb 26, 2008 4:37 amFollowed by: 1 member; Thanked: 5 times

by pranavc » Tue Jul 22, 2008 4:42 pm

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

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Source: Beat The GMAT — Problem Solving | Tue Jul 22, 2008 4:42 pm

gmatutor: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Sun Dec 17, 2006 1:53 pm; Location: London; Thanked: 7 times

by gmatutor » Tue Jul 22, 2008 4:55 pm

Suppose you drew two balls already. It doesn't matter what the sum of those two balls are.

If the sum is even then by drawing an odd on your last draw you will get an odd.

Likewise, if the sum is odd then by drawing an even on the last draw.

In this question only the last ball you draw that will determine if the sum is even or add. Since you have a 1/2 chance on your last draw of drawing an even or and odd, the probability that the sum will be odd is 1/2.

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pranavc: Master | Next Rank: 500 Posts; Posts: 167; Joined: Tue Feb 26, 2008 4:37 amFollowed by: 1 member; Thanked: 5 times

by pranavc » Tue Jul 22, 2008 4:58 pm

Thanks a ton for the prompt response. That is one slick explanation. Makes perfect sense.

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