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

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07 May 2012, 07:16
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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
[Reveal] Spoiler: OA

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

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07 May 2012, 07:46
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Galiya wrote:
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 $$C^2_2*C^3_7=35$$;

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

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

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07 May 2012, 07:51
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There are 3 spots left after selecting John and Peter in the team. There are 7 players available to fill in the 3 positions. It can be done in 7C3 ways (favorable event)
The total number of ways of selecting 5 out of 9 players is given by 9C5 ways.

7C3 / 9C5 = 5/18

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19 Aug 2012, 00:30
total combinations = 9C5 = 126
combinations with j & p = 1*1*\frac{7*6*5}{3!} = 35

probability = \frac{35}{126} = \frac{5}{18}
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Re: A basketball coach will select the members of a five-player [#permalink]

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20 Aug 2012, 09:22
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Galiya wrote:
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

Probability of choosing John and Peter for the team plus three other players is $$\frac{2}{9}*\frac{1}{8}*\frac{7}{7}*\frac{6}{6}*\frac{5}{5}=\frac{1}{36}.$$
Order choosing the two doesn't matter.
There is a total number of $$5C2=\frac{5*4}{2}=10$$ possibilities to choose the two (John and Peter) in the sequence of 5 players (like * * - - -, - * * - - ,... where * represents either John or Peter, no distinction between them).
The requested probability is given by $$10*\frac{1}{36}=\frac{5}{18}.$$

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

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23 Aug 2012, 09:58
Would there be an issue doing it this way?

Probability of John getting chosen = 5/9, probability of Peter being subsequently chosen = 4/8.
5/9 * 4/8 = 5/18

Why is this not the recommended approach given that the use of combinatorics requires extensive multiplication?

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

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23 Aug 2012, 10:32
dyaffe55 wrote:
Would there be an issue doing it this way?

Probability of John getting chosen = 5/9, probability of Peter being subsequently chosen = 4/8.
5/9 * 4/8 = 5/18

Why is this not the recommended approach given that the use of combinatorics requires extensive multiplication?

How can you justify the probability of John getting chosen = 5/9? John is just one player of the total of 9, so what's the meaning of 5?
Similar questions for Peter.
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Re: A basketball coach will select the members of a five-player [#permalink]

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23 Aug 2012, 12:08
My thinking was this... PA = (# of ways event can occur)/(total number of outcomes). John has 5 ways to be selected to the roster from a pool of 9 players = 5/9

perhaps I am missing something very obvious

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

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24 Aug 2012, 11:41
dyaffe55 wrote:
My thinking was this... PA = (# of ways event can occur)/(total number of outcomes). John has 5 ways to be selected to the roster from a pool of 9 players = 5/9

perhaps I am missing something very obvious

Mixing up things...

Number of outcomes is not 9, which is the total number of players to choose from.
Total number of outcomes is the number of different teams you can form choosing 5 players out of 9.
"John has 5 ways to be selected" has no meaning. You can count the number of possible teams of which John is also a member, but this is not 5.

Try to understand the posted solutions to this question.
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Re: A basketball coach will select the members of a five-player [#permalink]

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24 Aug 2012, 13:39
I'm not saying I don't understand the posted solutions... I do. However, I am attempting to find a valid alternative. All you keep saying is that "5 has no meaning" which really doesn't do much to refute my question.

"How can you justify the probability of John getting chosen = 5/9?"

If John is one of 9 and 5 random selections are made = 5/9.

However, I will agree that conceptually it isn't perfect after that. You could say the probability of peter getting selected next = 4/8. While I know this isn't exactly what the question asks, the result is the same. You could even add a third player and say "what is the probability that the coach chooses a team that includes John Peter and Paul." 5/9 x 4/8 x 3/7 = 6C2 / 9C5

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

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12 May 2014, 09:07
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Prob/perm approach. As each required player can be tied to either a 1/8 or 1/9 prob, permutation is required

(1/9)(1/8)(5P2)
=5/18
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Re: A basketball coach will select the members of a five-player [#permalink]

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06 Nov 2014, 17:18
Hi,

I solved it as:

Probability of picking random players plus picking John and Peter $$\frac{2}{9}*\frac{1}{8}*\frac{7}{7}*\frac{6}{6}*\frac{5}{5}=\frac{1}{36}.$$

Not sure where I went wrong?

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

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06 Nov 2014, 20:40
Probability approach:

2/9*1/8*1*1*1=2/72

symmetry is: 5!/2!*3!=10, so

2/72*10=20/72=5/18

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

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07 Nov 2014, 03:20
russ9 wrote:
Hi,

I solved it as:

Probability of picking random players plus picking John and Peter $$\frac{2}{9}*\frac{1}{8}*\frac{7}{7}*\frac{6}{6}*\frac{5}{5}=\frac{1}{36}.$$

Not sure where I went wrong?

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?

It should be P(JPAAA) = 1/9*1/8*7/7*6/6*5/5*5!/3! = 5/18. We multiply by 5!/3! because JPAAA can occur in several ways: JPAAA, JAPAAA, AJPAAA, ...
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Re: A basketball coach will select the members of a five-player [#permalink]

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20 Aug 2016, 03:49
well a much easier approach for me was
for Peter to get selected in the team : there is 5 empty spots out of 9 contenders so his chance 5/9
for john 4 spots left out of 8 so his chance 4/8
both chance= 4/8*5/9=5/18 easier than 9 choose 5 calculation

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11 Nov 2017, 21:52
Galiya wrote:
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

Number of combinations with constraint i.e., both John and Peter included in the team =7C3=35
Total number of combinations = 9C5=126
Probability= 35/126=5/18
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08 Dec 2017, 22:51
Think of the team as five slots left to right.
John then peter =(1/9)(1/8). We have to multiply that by 20 because they can occupy 20 different slot combinations. 20/72=10/36=5/18

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

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