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John and Peter are among the nine players a basketball coach [#permalink]

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21 Dec 2007, 07:26

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John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

Explain, please, your approach step by step. Your logic is not clear for me...How did you get 8/9, 7/8....? Maybe I missed something.

sure...

total number of 5 person teams, out of 9 players, possible: 9*8*7*6*5
number of teams that include John and Peter on the same team:
J P _ _ _ J and P counted as 1 person, 8*7*6*5 * 2 (multiply by two because JP and PJ placements)...

so:
number of Peter-John teams divided by total number of teams:

Explain, please, your approach step by step. Your logic is not clear for me...How did you get 8/9, 7/8....? Maybe I missed something.

sure...

total number of 5 person teams, out of 9 players, possible: 9*8*7*6*5 number of teams that include John and Peter on the same team: J P _ _ _ J and P counted as 1 person, 8*7*6*5 * 2 (multiply by two because JP and PJ placements)...

so: number of Peter-John teams divided by total number of teams:

8*7*6*5 * 2 / 9*8*7*6*5 = 2/9

hmm you know what--i just realized this should be a combination problem and not perm....

John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

1/9 1/6 2/9 5/18 1/3

I set it up like this 1/9*1/8 --> 1/72 then had 1/8*1/9 --> 1/72 --> 1/36... from here I just guessed D.

Now retrying this I think I can come up with a proper explanation.

how many different scenarios do we have John and Peter in a team of 5?

Should be 5!/(5!-2!) --> 5!/3! --> 20 ways in which we could arrange john and peter on a team of 5.

20/72--> 10/36--> 5/18.

Walker can u plz confirm whether approach is correct?

John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?

1/9 1/6 2/9 5/18 1/3

Total number of ways of choosing 5 people out of 9 = how many combinations are possible = 9C5.

Total number of favorite combinations, any of these combinations will always include John and Peter. thus we need to know how many combinations are feasible, when we have to choose 3 people out of 7 = 7C3.

Thus probability

P= 7C3/ 9C5 = 35/126 = 5/18.

Possible Wrong approach is as follows***********************

But initially, I had taken the wrong approach, and used the permutation fot the favorable events, I did something like this:

2 ( John and Peter) are already chosen, and now we need to choose the remaining 3 out of 7. Thus first can be chosen in 7 ways, second in 6 ways and third in 5 ways.

P= 7*6*5/ 9C5.

But 7*6*5 is permutation, just the number of ways the three people can be arranged and not the total number of combinations.

Just think/remember: ORDER does not matter in selecting the 3people out of 7, therefore it is a combination problem and not permutation ( in which order matters.)

I set it up like this 1/9*1/8 --> 1/72 then had 1/8*1/9 --> 1/72 --> 1/36... from here I just guessed D.

Looks like permutation...

GMATBLACKBELT wrote:

how many different scenarios do we have John and Peter in a team of 5?

Should be 5!/(5!-2!) --> 5!/3! --> 20 ways in which we could arrange john and peter on a team of 5.

20/72--> 10/36--> 5/18.

I guess your direction is not correct but you got right answer !!!
when you use 5!/3! it seems you should use the fact that there is the nine players but you did not use that....

I got this correct in 1 min. but interestingly when i tried different approach ans. was different..please tell me what I am doing wrong... total combinations 9c5 = 126

total combinations w/o those 2 players in the team (9 - 2)C5 => 7C5 = 21 probability that those 2 would not be selected 21/126 = 1/6 probability that those 2 would be selected 1-1/6 = 5/6 and guess what, it is not even an option <blushes>

what am i doing wrong?

gmatclubot

Re: PS probability
[#permalink]
17 Feb 2008, 12:00