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A basketball coach will select the members of a five-player [#permalink]
17 Aug 2010, 10:27

00:00

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Difficulty:

35% (medium)

Question Stats:

79% (02:16) correct
21% (02:55) wrong based on 29 sessions

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?

Re: MGMAT Probability [#permalink]
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

Re: MGMAT Probability [#permalink]
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?

Re: MGMAT Probability [#permalink]
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?

Re: MGMAT Probability [#permalink]
12 May 2014, 01:25

Expert's post

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;