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

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galiya
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A basketball coach will select the members of a five-player [#permalink] New post   07 May 2012, 08:16
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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
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D
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saishankari
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Re: A basketball coach will select the members of a five-player [#permalink] New post   07 May 2012, 08:51
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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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Re: A basketball coach will select the members of a five-player [#permalink] New post   20 Aug 2012, 10:22
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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 \(10*\frac{1}{36}=\frac{5}{18}.\)

Answer D
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Re: A basketball coach will select the members of a five-player [#permalink] New post   07 May 2012, 08:46
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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 \(C^2_2*C^3_7=35\);

\(P=\frac{35}{126}=\frac{5}{18}\).

Answer: D.
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Re: A basketball coach will select the members of a five-player [#permalink] New post   23 Aug 2012, 10:58
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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] New post   23 Aug 2012, 11:32
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] New post   23 Aug 2012, 13:08
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] New post   24 Aug 2012, 12:41
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] New post   12 May 2014, 10:07
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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] New post   06 Nov 2014, 18:18
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] New post   07 Nov 2014, 04:20
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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] New post   29 Jan 2018, 07:47
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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


P(John and Peter both on the team) = (# of teams that include both John and Peter) / (total # of 5-person teams possible)

a) # of teams that include both John and Peter
- Put John and Peter on the team. This can be accomplished in 1 way
- Select the remaining 3 team-members from the remaining 7 players. Since the order in which we select the 3 players does not matter, we can use combinations. We can select 3 players from 7 players in 7C3 ways (35 ways)
So, the total # of teams that include both John and Peter = (1)(35) = 35

b) total # of 5-person teams
Select 5 team-members from the 9 players. This can be accomplished in 9C5 ways
So, the total # of 5-person teams = 9C5 = 126

Therefore, the probability that the coach chooses a team that includes both John and Pete = 35/126 = 5/18

Answer: D

RELATED VIDEO (calculating combinations, like 7C3, in your head


Cheers,
Brent
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Re: A basketball coach will select the members of a five-player [#permalink] New post   05 Jul 2018, 16:51
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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


The number of ways of choosing 5 players from 9 is 9C5 = 9!/[5!(9 - 5)!] = 9!/(5!4!) = (9 x 8 x 7 x 6 )/(4 x 3 x 2) = 3 x 7 x 6 = 126.

If John and Peter are already chosen for the team, then only 3 additional players must be chosen for the five-player team. The number of ways of choosing the 3 additional players from the remaining 7 players is 7C3 = 7!/[3!(7 - 3)!] = 7!/(3!4!) = (7 x 6 x 5)/(3 x 2) = 7 x 5 = 35.

Thus, the probability that John and Peter will be chosen for the team is 35/126 = 5/18.

Answer: D
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Re: A basketball coach will select the members of a five-player [#permalink] New post   16 Mar 2021, 10:27
BrentGMATPrepNow wrote:
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


P(John and Peter both on the team) = (# of teams that include both John and Peter) / (total # of 5-person teams possible)

a) # of teams that include both John and Peter
- Put John and Peter on the team. This can be accomplished in 1 way
- Select the remaining 3 team-members from the remaining 7 players. Since the order in which we select the 3 players does not matter, we can use combinations. We can select 3 players from 7 players in 7C3 ways (35 ways)
So, the total # of teams that include both John and Peter = (1)(35) = 35

b) total # of 5-person teams
Select 5 team-members from the 9 players. This can be accomplished in 9C5 ways
So, the total # of 5-person teams = 9C5 = 126

Therefore, the probability that the coach chooses a team that includes both John and Pete = 35/126 = 5/18

Answer: D

RELATED VIDEO (calculating combinations, like 7C3, in your head


Cheers,
Brent


BrentGMATPrepNow i dont get highlighted part, why it can be done in 1 way ? choosing P and J

it can be pxxxj, or pxxjx etc so many ways to choose
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Re: A basketball coach will select the members of a five-player [#permalink] New post   16 Mar 2021, 11:10
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dave13 wrote:
BrentGMATPrepNow wrote:
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


P(John and Peter both on the team) = (# of teams that include both John and Peter) / (total # of 5-person teams possible)

a) # of teams that include both John and Peter
- Put John and Peter on the team. This can be accomplished in 1 way
- Select the remaining 3 team-members from the remaining 7 players. Since the order in which we select the 3 players does not matter, we can use combinations. We can select 3 players from 7 players in 7C3 ways (35 ways)
So, the total # of teams that include both John and Peter = (1)(35) = 35

b) total # of 5-person teams
Select 5 team-members from the 9 players. This can be accomplished in 9C5 ways
So, the total # of 5-person teams = 9C5 = 126

Therefore, the probability that the coach chooses a team that includes both John and Pete = 35/126 = 5/18

Answer: D

RELATED VIDEO (calculating combinations, like 7C3, in your head


Cheers,
Brent


BrentGMATPrepNow i dont get highlighted part, why it can be done in 1 way ? choosing P and J

it can be pxxxj, or pxxjx etc so many ways to choose

The assumption here is that order doesn't matter. So pxxxj and pxxjx would be considered the same outcome.
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Re: A basketball coach will select the members of a five-player [#permalink] New post   18 Jul 2022, 14:26
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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


There are 9 slots to fill. 5 of them are on the team; 4 of them are not on the team.
There is a 5/9 probability of any individual player being on the team (including either John or Peter). Once any individual player is on the team, there is a 4/8 probability of any remaining player being on the team.
5/9 * 4/8 = 20/72 = 5/18

Answer choice D.
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Re: A basketball coach will select the members of a five-player [#permalink] New post   26 Jul 2023, 22:21
Probability = No. of favourable outcomes / No. of total outcomes
= 7C3/9C5
=35/126
=5/18

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