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25 Jul 2010, 01:50
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
75% (02:03) correct
25%
(02:35)
wrong
based on 175
sessions
History
Date
Time
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Not Attempted Yet
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
a. 1/9
b. 1/6
c. 2/9
d. 5/18
e. 1/3
(2C2 x 7C3)/9C5=5/18
But when i do (1- Neg of prob) , I am getting incorrect answer.
P(A and B not selected)= 7c5/9c5
P(A and B Selected)=1-7c5/9c5=1=1-1/6=5/6
what am i doing wrong
But when i do (1- Neg of prob) , I am getting incorrect answer.
P(A and B not selected)= 7c5/9c5
P(A and B Selected)=1-7c5/9c5=1=1-1/6=5/6
what am i doing wrong
25 Jul 2010, 02:26
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anaik100
Finding the probability of some event by finding the probability of opposite event and subtracting it from 1 is a good strategy, sometimes the best. But this approach also has some traps and most common trap is when one defines an opposite event incorrectly.
In original question we are asked to find the probability of an event that both A and B will be in the team, the opposite event would be: the event that none of them is in the team PLUS event that only one of them is in the team (you calculated the probability of just the first part of the opposite event).
Probability that none of them is in the team is \(\frac{C^5_7}{C^5_9}=\frac{1}{6}\);
Probability that only one will be in the team is \(2*\frac{C^1_1*C^4_7}{C^5_9}=\frac{5}{9}\);
So, \(P=1-(\frac{1}{6}+\frac{5}{9})=\frac{5}{18}\).
You can see in this case it's better to calculate probability of event directly, as you did in your first approach - \(P=\frac{C^2_2*C^3_7}{C^5_9}=\frac{5}{18}\).
Answer: D.
Hope it helps.
30 Jan 2018, 11:25
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anaik100
The number of ways to select the team with no restrictions is 9C5 = 9!/[5!(9-5)!] = 9!/(5!4!) = (9 x 8 x 7 x 6 x 5)/5! =(9 x 8 x 7 x 6 x 5)/(5 x 4 x 3 x 2 x 1) = 126 ways
Since we must have A and B on the team, then there are only 7 players remaining to be chosen for the team, and there will be only 3 spots on the team for them to fill. The number of ways to select the team, which already has A and B selected, is 7C3 = 7!/3![7-3)!] = 7!/(3!4!) = (7 x 6 x 5)/3! = (7 x 6 x 5)/(3 x 2 x 1) = 35 ways
Thus, the the probability that the coach chooses a team that includes both A and B is 35/126 = 5/18.
Answer: D
