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# A coach will select the members of a 5-players team from 9 players, in

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A coach will select the members of a 5-players team from 9 players, in  [#permalink]

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25 Jul 2010, 01:50
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Difficulty:

35% (medium)

Question Stats:

74% (02:00) correct 26% (02:19) wrong based on 82 sessions

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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
(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

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Posts: 50041
Re: A coach will select the members of a 5-players team from 9 players, in  [#permalink]

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25 Jul 2010, 02:26
1
anaik100 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

(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

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 - $$P=\frac{C^2_2*C^3_7}{C^5_9}=\frac{5}{18}$$.

Hope it helps.
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Re: A coach will select the members of a 5-players team from 9 players, in  [#permalink]

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30 Jan 2018, 11:25
2
anaik100 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

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.

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Re: A coach will select the members of a 5-players team from 9 players, in &nbs [#permalink] 30 Jan 2018, 11:25
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