Combinatrics/Probability Q

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Strongt: Senior | Next Rank: 100 Posts; Posts: 51; Joined: Wed Jan 05, 2011 3:32 amFollowed by: 1 member

by Strongt » Wed May 25, 2011 6:06 am

8 students have been chosen to play for PCU's inter"collegiate
basketball team. If every person on the team has an equal chance of
starting, what is the probability that both Tom and Alex will start?
(Assume 5 starting positions)

Answer is [spoiler]5/14[/spoiler]

source MGMAT flashcards

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Source: Beat The GMAT — Problem Solving | Wed May 25, 2011 6:06 am

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 » Wed May 25, 2011 6:18 am

You could reason that for them both to start, neither of them can be among the 3 players who will not start: The probability that the three not starting do not include either is 6/8 x 5/7 x 4/6 = 5/14

Alternatively, if you are comfortable with the combinations formula, you could think that there are 8C5 =8C3 = 56 subsets of size 3, all of which are equally probable; 2C2 x 6C3 = 20 of these contain both of the players mentioned: 20/56 = 5/14

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