Vincen wrote:A coach will select the members of a 5-players team from 9 players, including A and B. If the 5 players are chosen at random, What is the probability that the coach chooses a team that includes both A and B?
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(A is included) = 5/9.
Since 4 of the remaining 8 players are included on the team, P(B 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 A and B)/(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 A and B:
Once A and B have been selected, the coach must select 3 additional players to combine with A and B.
The result will be a 5-member team with A and B.
From the 7 remaining players, the number of ways to choose 3 to combine with A and B = 7C3 = (7*6*5)/(3*2*1) = 35.
Resulting probability:
(5-member teams with A and B)/(all possible 5-member teams) = 35/126 = 5/18.
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