BTGmoderatorLU wrote: ↑

Thu Mar 05, 2020 4:03 pm

Source: Manhattan Prep

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

The OA is D

We want:
a) # of teams that include both John and Peter
b) total # of 5-person 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 team-members 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 5-person teams
Select 5 team-members from the 9 players (this can be accomplished in 9C5 ways)
So, the total # of 5-person teams = 9C5 = 126

Therefore, the probability that the coach chooses a team that includes both John and Pete = 35/126 = 5/18

Answer: D

Cheers,
Brent

BTGmoderatorLU wrote: ↑

Thu Mar 05, 2020 4:03 pm

Source: Manhattan Prep

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

The OA is D

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.

If John and Peter are already chosen for the team, then only 3 additional players must be chosen for the five-player 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