sparkles3144 wrote: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!) b) (6*5*4)/(8*7*6*5*4) c) (6!/3!3!)/(8*7*6*5*4)
Of the 8 students, 5 are to be chosen to start.
P(Tom is chosen) = 5/8.
Of the 7 remaining students, 4 are to be chosen to start.
P(Alex is chosen) = 4/7.
Since we want both events to happen, we MULTIPLY the fractions:
5/8 * 4/7 = 5/14.
A: ( 6!/3!3! )/ ( 8!/5!3! ) = (5*4)/(8*7) = 5/14.
The correct answer is
A.
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