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

Question

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

A. 1/9
B. 1/6
C. 2/9
D. 5/18
E. 1/3

saishankari · Accepted Answer

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

EvaJager · Answer

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

Bunuel · Answer

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.

dyaffe55 · Answer

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?

EvaJager · Answer

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.

dyaffe55 · Answer

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

EvaJager · Answer

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.

m3equals333 · Answer

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

russ9 · Answer

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?

Bunuel · Answer

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

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, ...

BrentGMATPrepNow · Answer

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
 https://www.youtube.com/watch?v=05a0A7vjG8I

Cheers, 
Brent

ScottTargetTestPrep · Answer

The number of ways of choosing 5 players from 9 is 9C5 = 9!/  = 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!/  = 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

dave13 · Answer

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

BrentGMATPrepNow · Answer

The assumption here is that order doesn't matter. So pxxxj and pxxjx would be considered the same outcome.

ThatDudeKnows · Answer

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.

Paras96 · Answer

Probability = No. of favourable outcomes / No. of total outcomes 
= 7C3/9C5
=35/126
=5/18

Hence D

FelixLucas · Answer

OMG, finally an answer that is not ridiculously long....... I mean.... Anyway, Thanks a lot.

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

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

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