John and Peter are among the nine players a basketball coach can choose from to field a fiveplayer team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?
A 1/9
B 1/6
C 2/9
D 5/18
E 1/3
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John and Peter are among the nine players a basketball coach can choose from to field a fiveplayer team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?
A 1/9
B 1/6
C 2/9
D 5/18
E 1/3
Number of ways of making a free selection of a team of 5 players from 9 players = 9C5 = 9*8*7*6/4*3*2*1 = 9*7*2 = 63*2 = 126
Number of ways of selecting 3 players from amongst 7 players when John and Peter have already been selected = 2C2 * 7C3 = 1 * [7*6*5 / 3*2*1] = 35
Thus probability = 35/126 = 5/18
Hence D
A 1/9
B 1/6
C 2/9
D 5/18
E 1/3
Number of ways of making a free selection of a team of 5 players from 9 players = 9C5 = 9*8*7*6/4*3*2*1 = 9*7*2 = 63*2 = 126
Number of ways of selecting 3 players from amongst 7 players when John and Peter have already been selected = 2C2 * 7C3 = 1 * [7*6*5 / 3*2*1] = 35
Thus probability = 35/126 = 5/18
Hence D
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 Scott@TargetTestPrep
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The number of ways of choosing 5 players from 9 is 9C5 = 9!/[5!(9  5)!] = 9!/(5!4!) = (9 x 8 x 7 x 6 )/( 4 x 3 x 2) = 3 x 7 x 6 = 126.Tame the CAT wrote:John and Peter are among the nine players a basketball coach can choose from to field a fiveplayer team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?
A 1/9
B 1/6
C 2/9
D 5/18
E 1/3
If John and Peter are already chosen for the team, then only 3 additional players can be chosen for the fiveplayer team. The number of ways of choosing the 3 additional players from the remaining 7 players is 7C3 = 7!/[3!(7  3)!] = 7!/(3!4!) = (7 x 6 x 5)/(3 x 2) = 7 x 5 = 35.
Thus, the probability that John and Peter will be chosen for the team is 35/126 = 5/18.
Answer: D
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 Brent@GMATPrepNow
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We want:Tame the CAT wrote:John and Peter are among the nine players a basketball coach can choose from to field a fiveplayer team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?
A 1/9
B 1/6
C 2/9
D 5/18
E 1/3
a) # of teams that include both John and Peter
b) total # of 5person teams possible
a) # of teams that include both John and Peter
Put John and Peter on the team (this can be accomplished in 1 way)
Select the remaining 3 teammembers from the remaining 7 players (this can be accomplished in 7C3 ways)
So, the total # of teams that include both John and Peter = (1)(7C3) = 35
b) total # of 5person teams
Select 5 teammembers from the 9 players (this can be accomplished in 9C5 ways)
So, the total # of 5person teams = 9C5 = 126
Therefore, the probability that the coach chooses a team that includes both John and Pete = 35/126 = 5/18
Answer: D
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