## Tricky probability

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### Tricky probability

by sparkle6 » Thu Sep 29, 2011 6:52 am
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

[spoiler]Can somebody please explain using combination theory?[/spoiler]

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by [email protected] » Thu Sep 29, 2011 6:59 am
sparkle6 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
Can somebody please explain using combination theory?
The combination theory approach:

We want:
a) # of teams that include both John and Peter
b) 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 (this can be accomplished in 7C3 ways)
So, the total # of teams that include both John and Peter = (1)(7C3) = 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 = D

Cheers,
Brent
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by [email protected] » Thu Sep 29, 2011 7:04 am
By the way, if you'd like to learn how to quickly calculate combinations (such as 7C3) in your head, you can watch video #17 at https://www.gmatprepnow.com/module/gmat-counting (it's free)

Cheers,
Brent
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by gmattesttaker2 » Sun Jul 08, 2012 11:45 pm
[email protected] wrote:
sparkle6 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
Can somebody please explain using combination theory?
The combination theory approach:

We want:
a) # of teams that include both John and Peter
b) 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 (this can be accomplished in 7C3 ways)
So, the total # of teams that include both John and Peter = (1)(7C3) = 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 = D

Cheers,
Brent

Hello Brent,

I was just wondering if we can solve this problem using the slot method like how you have explained here:

https://www.beatthegmat.com/probability- ... tml#481625

I tried something similar here:

P(John Peter Player1 Player2 Player3) = P(John on 1st) x P(Peter on 2nd) x P(Player1 on 3rd) x P(Player2 on 4th) x P(Player3 on 4th)

= (1/9).(1/8).(7/7).(6/6).(5/5)
= 1/72

After this I tried to arrange the 5 players as follows: 5 x 4 x 3 x 2 x 1 = 120
and then multiply 120 x (1/72)

However, I am going wrong since the answer is not 5/18. I was wondering if you can please assist here? Thanks for all your valuable time and help.

Best Regards,
Sri

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by [email protected] » Mon Jul 09, 2012 6:28 am
gmattesttaker2 wrote:
[email protected] wrote:
sparkle6 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
Can somebody please explain using combination theory?
The combination theory approach:

We want:
a) # of teams that include both John and Peter
b) 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 (this can be accomplished in 7C3 ways)
So, the total # of teams that include both John and Peter = (1)(7C3) = 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 = D

Cheers,
Brent

Hello Brent,

I was just wondering if we can solve this problem using the slot method like how you have explained here:

https://www.beatthegmat.com/probability- ... tml#481625

I tried something similar here:

P(John Peter Player1 Player2 Player3) = P(John on 1st) x P(Peter on 2nd) x P(Player1 on 3rd) x P(Player2 on 4th) x P(Player3 on 4th)

= (1/9).(1/8).(7/7).(6/6).(5/5)
= 1/72

After this I tried to arrange the 5 players as follows: 5 x 4 x 3 x 2 x 1 = 120
and then multiply 120 x (1/72)

However, I am going wrong since the answer is not 5/18. I was wondering if you can please assist here? Thanks for all your valuable time and help.

Best Regards,
Sri

Rather that think of it as P(John Peter Player1 Player2 Player3) as you have in your solution, think of it as P(John Peter Other Other Other)
Your calculation is fine. P(John Peter Other Other Other) = 72
Here we need to recognized that John Peter Other Other Other is just one acceptable outcome.
Other acceptable outcomes include
John Other Peter Other Other
Other Other John Peter Other
Other John Other Peter Other
.
.
.

In how many ways can we arrange the 5 objects (where 3 are identical)? We need to apply the MISSISSIPPI rule (as I explained in https://www.beatthegmat.com/probability- ... tml#481625)

We can do this in 5!/3! ways (20 ways)

So, the desired probability = (1/72)(20) = 5/18

Cheers,
Brent
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by gmattesttaker2 » Mon Jul 09, 2012 11:13 pm
[email protected] wrote:
gmattesttaker2 wrote:
[email protected] wrote:
sparkle6 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
Can somebody please explain using combination theory?
The combination theory approach:

We want:
a) # of teams that include both John and Peter
b) 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 (this can be accomplished in 7C3 ways)
So, the total # of teams that include both John and Peter = (1)(7C3) = 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 = D

Cheers,
Brent

Hello Brent,

I was just wondering if we can solve this problem using the slot method like how you have explained here:

https://www.beatthegmat.com/probability- ... tml#481625

I tried something similar here:

P(John Peter Player1 Player2 Player3) = P(John on 1st) x P(Peter on 2nd) x P(Player1 on 3rd) x P(Player2 on 4th) x P(Player3 on 4th)

= (1/9).(1/8).(7/7).(6/6).(5/5)
= 1/72

After this I tried to arrange the 5 players as follows: 5 x 4 x 3 x 2 x 1 = 120
and then multiply 120 x (1/72)

However, I am going wrong since the answer is not 5/18. I was wondering if you can please assist here? Thanks for all your valuable time and help.

Best Regards,
Sri

Rather that think of it as P(John Peter Player1 Player2 Player3) as you have in your solution, think of it as P(John Peter Other Other Other)
Your calculation is fine. P(John Peter Other Other Other) = 72
Here we need to recognized that John Peter Other Other Other is just one acceptable outcome.
Other acceptable outcomes include
John Other Peter Other Other
Other Other John Peter Other
Other John Other Peter Other
.
.
.

In how many ways can we arrange the 5 objects (where 3 are identical)? We need to apply the MISSISSIPPI rule (as I explained in https://www.beatthegmat.com/probability- ... tml#481625)

We can do this in 5!/3! ways (20 ways)

So, the desired probability = (1/72)(20) = 5/18

Cheers,
Brent

Hello Brent,

Many, many thanks for your prompt reply and for the excellent explanation (as always!). It is clear now where I was going wrong. You explain everything so clearly and systematically. Thank you very much again.

Best Regards,
Sri

### GMAT/MBA Expert

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Posts: 15041
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by [email protected] » Tue Jul 10, 2012 6:44 am
gmattesttaker2 wrote:Hello Brent,

Many, many thanks for your prompt reply and for the excellent explanation (as always!). It is clear now where I was going wrong. You explain everything so clearly and systematically. Thank you very much again.

Best Regards,
Sri