A basketball coach will select the members of a five-player

[BTGmoderatorLU]

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.

Please, can anyone assist me with this PS question? I don't understand it and I need any suggestion on how to solve it. Thanks!

[Keith@ThePrincetonReview]

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.

Please, can anyone assist me with this PS question? I don't understand it and I need any suggestion on how to solve it. Thanks!

The question asks for a probability, so map out the answer by setting up a fraction. The numerator represents the number of teams that include both Peter and John, and the denominator represents the total number of possible teams.

Total number of possible teams:
Imagine that the coach is selecting the five-player team one player at a time. There are 9 players that could be selected first, 8 that could be selected second, 7 that could be selected third, and so on. The number of ways the coach could select the 5 members on the team is therefore 9 × 8 × 7 × 6 × 5. However, some of those teams will consist of the same 5 players. To eliminate duplicate teams, divide by 5! (because the coach is choosing 5 players to be on the team). The total number of unique teams is (9×8×7×6×5)/(5×4×3×2×1)=126.

Number of teams that include both Peter and John:
Because Peter and John must be on the team, the coach will choose 3 of the remaining 7 players to be on the team. Repeat the steps above using these values, so that the number of teams that include both Peter and John is (7×6×5)/(3×2×1)=35.

The probability that the coach choose a team that includes both Peter and John is 35/126, which is equal to 5/18.

Answer: D

[GMATGuruNY]

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.

[Brent@GMATPrepNow]

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

P(John and Peter both on the team) = (# of teams that include both John and Peter) / (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. Since the order in which we select the 3 players does not matter, we can use combinations. We can select 3 players from 7 players in 7C3 ways (35 ways)
So, the total # of teams that include both John and Peter = (1)(35) = 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

Aside: If anyone is interested, we have a free video on calculating combinations (like 7C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Cheers,
Brent

[Scott@TargetTestPrep]

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 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