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

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Current Student
Joined: 31 Aug 2007
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John and Peter are among the nine players a basketball coach [#permalink]

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21 Dec 2007, 08:26
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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?

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

Kudos [?]: 163 [0], given: 1

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4587 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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21 Dec 2007, 09:05
D

P=(9-2)C(5-2)*2C2/9C5=7!*2!*5!*4!/(3!*4!*9!*2!)=5*4/(8*9)=5/18

Kudos [?]: 4587 [0], given: 360

Current Student
Joined: 31 Aug 2007
Posts: 368

Kudos [?]: 163 [0], given: 1

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21 Dec 2007, 09:45
walker wrote:
D

P=(9-2)C(5-2)*2C2/9C5=7!*2!*5!*4!/(3!*4!*9!*2!)=5*4/(8*9)=5/18

hmm walker why isn't this correct:

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

# of teams with the two required people divided by total number of teams

Kudos [?]: 163 [0], given: 1

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4587 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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21 Dec 2007, 10:56
young_gun wrote:
hmm walker why isn't this correct:
8*7*6*5*2 / 9*8*7*6*5 = 2/9

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.

Kudos [?]: 4587 [0], given: 360

Current Student
Joined: 31 Aug 2007
Posts: 368

Kudos [?]: 163 [0], given: 1

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21 Dec 2007, 11:02
walker wrote:
young_gun wrote:
hmm walker why isn't this correct:
8*7*6*5*2 / 9*8*7*6*5 = 2/9

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

Kudos [?]: 163 [0], given: 1

Current Student
Joined: 31 Aug 2007
Posts: 368

Kudos [?]: 163 [0], given: 1

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21 Dec 2007, 11:03
young_gun wrote:
walker wrote:
young_gun wrote:
hmm walker why isn't this correct:
8*7*6*5*2 / 9*8*7*6*5 = 2/9

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....

Kudos [?]: 163 [0], given: 1

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4587 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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21 Dec 2007, 11:12
young_gun wrote:
hmm you know what--i just realized this should be a combination problem and not perm....

Exactly! I also went by wrong way using permutation at the begin, but we really choose persons, don't positions.

Kudos [?]: 4587 [0], given: 360

CEO
Joined: 29 Mar 2007
Posts: 2554

Kudos [?]: 516 [0], given: 0

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21 Dec 2007, 15:36
young_gun wrote:
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?

Kudos [?]: 516 [0], given: 0

Director
Joined: 03 Sep 2006
Posts: 865

Kudos [?]: 1072 [0], given: 33

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21 Dec 2007, 20:27
young_gun wrote:
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.)

Kudos [?]: 1072 [0], given: 33

SVP
Joined: 28 Dec 2005
Posts: 1545

Kudos [?]: 179 [0], given: 2

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21 Dec 2007, 20:49
what about approach of 1-prob that both are not on the team ?

prob that they are both not on the team is what im having trouble figuring out though ...

Kudos [?]: 179 [0], given: 2

Director
Joined: 03 Sep 2006
Posts: 865

Kudos [?]: 1072 [0], given: 33

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21 Dec 2007, 21:09
pmenon wrote:
what about approach of 1-prob that both are not on the team ?

prob that they are both not on the team is what im having trouble figuring out though ...

When both are not in the team answer would be

P = 1-(7C3/9C5) = 1-(5/18)

Kudos [?]: 1072 [0], given: 33

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4587 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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22 Dec 2007, 01:47
GMATBLACKBELT wrote:
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....

Kudos [?]: 4587 [0], given: 360

CEO
Joined: 21 Jan 2007
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Kudos [?]: 1049 [0], given: 4

Location: New York City

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14 Feb 2008, 09:13
total = 9c5
teams with peter and tom = 7c3

desired/total = 7c3/9c5
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

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Senior Manager
Joined: 18 Jun 2007
Posts: 286

Kudos [?]: 61 [0], given: 0

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17 Feb 2008, 13:00
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?

Kudos [?]: 61 [0], given: 0

Re: PS probability   [#permalink] 17 Feb 2008, 13:00
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