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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