GMATPrep Probability Q

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ankanas: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Thu May 24, 2007 3:44 pm

by ankanas » Thu Aug 30, 2007 10:59 pm

How do i solve this one ?

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Source: Beat The GMAT — Problem Solving | Thu Aug 30, 2007 10:59 pm

agps: Master | Next Rank: 500 Posts; Posts: 101; Joined: Tue Aug 07, 2007 1:45 amFollowed by: 1 member

by agps » Fri Aug 31, 2007 1:16 am

total possibilities = 27 (3*3*3 because he can ride any of the 3 cars each time)
favorable possibilities = 6 (1-2-3, 1-3-2, 2-1-3, 2-3-1, 3-1-2, 3-2-1)

6/27 = 2/9

Answer C

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kevincanspain: GMAT Instructor; Posts: 613; Joined: Thu Mar 22, 2007 6:17 amFollowed by: 64 members; Location: madrid; Thanked: 171 times; GMAT Score:790

by kevincanspain » Fri Aug 31, 2007 2:17 pm

This is the probability that the second ride is not in the same car as the first multiplied by the probability that the third ride is in the car not previously ridden in given that the first and second rides were in different cars.

2/3 * 1/3 =2/9

Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid

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