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by parulmahajan89 » Mon Jan 20, 2014 8:58 pm
Which is the correct expression for this problem?
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)

a)
6!
3!3!
8!
5!3!

Can someone help with explanation of this answer?
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by GMATGuruNY » Tue Jan 21, 2014 12:07 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)
Of the 8 students, 5 will be starters, 3 will be non-starters.
Tom and Alex will both be starters if neither is selected to be a non-starter.
P(1st non-starter is not Tom or Alex) = 6/8. (Of the 8 students, there are 6 besides Tom and Alex.)
P(2nd non-starter is not Tom or Alex) = 5/7. (Of the 7 remaining students, there are 5 besides Tom and Alex.)
P(3rd non-starter is not Tom or Alex) = 4/6. (Of the 6 remaining students, there are 4 besides Tom and Alex.)
Since we want all 3 events to happen, we multiply the fractions:
6/8 * 5/7 * 4/6 = 5/14.
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by [email protected] » Tue Jan 21, 2014 1:03 pm
Hi parulmahajan89,

Mitch has properly explained how to solve the probability question that you posted. The answer choices that you provided don't match the type of question that's asked (by definition, any probability will be less than or equal to 1; in almost all cases, the probability will be a fraction).

If a question asked for the number of possible starting lineups, then you would likely use factorials.

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