BTGmoderatorLU wrote:A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter?
A. 1/9
B. 1/6
C. 2/9
D. 5/18
E. 1/3
Since 5 of the 9 players are included on the team, P(John is included) = 5/9.
Since 4 of the remaining 8 players are included on the team, P(Peter is included) = 4/8.
To combine the probabilities, we multiply:
5/9 * 4/8 = 5/18.
The correct answer is
D.
Alternate approach:
P = (5-member teams with John and Peter)/(all possible 5-member teams).
All possible 5-member teams:
From the 9 players, the number of ways to choose 5 = 9C5 = (9*8*7*6*5)/(5*4*3*2*1) = 126.
5-member teams with John and Peter:
Once John and Peter have been selected, the coach must select 3 additional players to combine with them.
The result will be a 5-member team with John and Peter.
From the 7 remaining players, the number of ways to choose 3 to combine with John and Peter = 7C3 = (7*6*5)/(3*2*1) = 35.
Resulting probability:
(5-member teams with John and Peter)/(all possible 5-member teams) = 35/126 = 5/18.
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