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# A basketball coach will select the members of a five-player

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17 Aug 2010, 10:27
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69% (01:06) correct 31% (01:04) wrong based on 75 sessions

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A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter?

A. 1/9
B. 1/6
C. 2/9
D. 5/18
E. 1/3

OPEN DISCUSSION OF THIS TOPIC IS HERE: a-basketball-coach-will-select-the-members-of-a-five-player-132143.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 May 2014, 23:10, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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17 Aug 2010, 13:11
you can but that will require more cases and will be time consuming.

straight forward way is to group J and P as always present on the team and as order does not matter so we just need to find the total number of ways to select rest three players = 7c3

total number of ways of selecting 5 players out of 9 = 9c5

probability = 7c3/9c5 = 5/18

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17 Aug 2010, 22:22
ingoditrust wrote:
you can but that will require more cases and will be time consuming.

probability = 7c3/9c5 = 5/18

Look at my approach - it's easy, nice and not at all time consuming. But I'm somehow getting the wrong answer
I decided to find the probability of not getting these two guys in the choosen group and then substruct it from 1.
What's wrong with my approach?

1-(7/9)*(6/8)*(5/7)*(4/6)*(3/5)= 15/18

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17 Aug 2010, 23:30
Financier wrote:
I decided to find the probability of not getting these two guys in the choosen group and then substruct it from 1.
What's wrong with my approach?

Prob of choosing both J and P = 1 - ( Prob of choosing neither J nor P + Prob of choosing J and not P + Prob of choosing P and not J)

Hope this helps. This was highlighted in one of the other probability threads earlier this day only.

Thanks.
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11 May 2014, 16:30
I'm not sure if I'm misunderstanding the question but the way I read it, the question states that "what is the probability that jon and peter" will be selected. Correct?

If that's the case, why isn't this method working:

(5c2)/(5c9) - I need to pick J and P from a team of 5 and the number of options for the teams are 5 out of 9 players. Why is that wrong?

Additionally, I tried the direct probability approach and that seemed to work out:

(2/9)(1/8)(5!/2!3!) - This implies that the prob of picking J or P is 2 out of 9, then only J or P is left so 1 out of 9 and then I could have picked them the 1st time, 2nd, 3rd etc...

This method seems to work but I actually have a question regarding the permutation part. It obviously doesn't matter if I pick John 1st, 2nd, 3rd etc...and the same goes for peter. If order doesn't matter, then why do I need to apply the permutation part of this?

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12 May 2014, 01:25
russ9 wrote:
I'm not sure if I'm misunderstanding the question but the way I read it, the question states that "what is the probability that jon and peter" will be selected. Correct?

If that's the case, why isn't this method working:

(5c2)/(5c9) - I need to pick J and P from a team of 5 and the number of options for the teams are 5 out of 9 players. Why is that wrong?

Additionally, I tried the direct probability approach and that seemed to work out:

(2/9)(1/8)(5!/2!3!) - This implies that the prob of picking J or P is 2 out of 9, then only J or P is left so 1 out of 9 and then I could have picked them the 1st time, 2nd, 3rd etc...

This method seems to work but I actually have a question regarding the permutation part. It obviously doesn't matter if I pick John 1st, 2nd, 3rd etc...and the same goes for peter. If order doesn't matter, then why do I need to apply the permutation part of this?

About 5C2 in the numerator: are we choosing 2 from 5? No, we want to choose 5 players, including John and Peter, from 9 players.

About the probability approach: though the order of the selection does not matter, John and Peter can be any from the 5 selected. We can have {John, Peter, any, any, any}, {John, any, Peter, any, any}, {John, any, any, Peter, any}, ... Now, each of these cases have equal probabilities to occur: 1/9*1/8*1*1*1 and there can be 5!/3! cases, so the overall probability is 1/9*1/8*1*1*1*5!/3! = 5/18.

A basketball coach will select the members of a five-player team from among 9 players, including John and Peter. If the five players are chosen at random, what is the probability that the coach chooses a team that includes both John and Peter?

A. 1/9
B. 1/6
C. 2/9
D. 5/18
E. 1/3

Total # of five-player teams possible is $$C^5_9=126$$;
# of teams that include both John and Peter is $$C^2_2*C^3_7=35$$;

$$P=\frac{35}{126}=\frac{5}{18}$$.

OPEN DISCUSSION OF THIS TOPIC IS HERE: a-basketball-coach-will-select-the-members-of-a-five-player-132143.html
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Re: MGMAT Probability   [#permalink] 12 May 2014, 01:25
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