John and Peter are among the nine players a basketball...

swerve wrote:John and Peter are among the nine players a basketball coach can choose from to field a five-player 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

The OA is D.

Method 1:

Team with John and Peter = Total ways of choosing team - Teams without John and Peter - Team with either of them = 9C5 - 7C5 - 2*7C4 = 126 - 21 - 70 = 35
Total ways of choosing team = 9C5 = 126
Probability = 35/126 = 5/18

Method 2:

Since John and peter are already in team so total ways of choosing rest of the team = 7C3 = 35
Total ways of choosing team = 9C5 = 126
Probability = 35/126 = 5/18

Answer: option D

I hope this helps!!!

swerve wrote:John and Peter are among the nine players a basketball coach can choose from to field a five-player 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

One approach:

From the 9 players, 5 are to be selected for the team.
P(john is selected) = 5/9.
From the 8 remaining players, 4 are to be selected for the team.
P(Peter is selected) = 4/8.
Multiplying the probabilities, we get:
5/9 * 4/8 = 5/18.

The correct answer is D.

An alternate approach is to determine the value of following fraction:
(good teams)/(all possible teams)

All possible teams:
From the 9 players, the number of ways to choose 5 = 9C5 = (9*8*7*6*5)/(5*4*3*2*1) = 126.

Good teams:
A good team will consist of John and Peter combined with 3 other players.
Since John and Peter must be included in a good team, our only concern is the number of ways we can choose 3 OTHER PLAYERS to combine with John and Peter.
From the 7 players besides John and Peter, the number of ways to choose 3 = 7C3 = (7*6*5)/(1*2*3) = 35.

Thus:
(good teams)/(all possible teams) = 35/126 = 5/18.

This is what I tried,

find out the probability of not choosing John and Peter

7/9 * 6/8 * 5/7 * 4/6 * 3/5 = 1/6

1 - 1/6 = 5/6

where did I go wrong?

P(team with both John and Peter) = 1 - P(team with neither John nor Peter) - P(team with John but not Peter) - P(team with Peter but not John).
To determine the desired probability, all of the colored probabilities above must be subtracted from 1.
Your solution neglects to subtract the two probabilities in red.

swerve wrote:John and Peter are among the nine players a basketball coach can choose from to field a five-player 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

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