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.
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