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

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17 Aug 2010, 11:27

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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?

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

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?

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?

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 \(C^2_2*C^3_7=35\);

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