A basketball coach will select the members of a fiveplayer 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]
Tricky probability
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The combination theory approach:sparkle6 wrote:A basketball coach will select the members of a fiveplayer 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?
We want:
a) # of teams that include both John and Peter
b) total # of 5person 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 teammembers 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 5person teams
Select 5 teammembers from the 9 players (this can be accomplished in 9C5 ways)
So, the total # of 5person 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 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/gmatcounting (it's free)
Cheers,
Brent
Cheers,
Brent

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[email protected] wrote:The combination theory approach:sparkle6 wrote:A basketball coach will select the members of a fiveplayer 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?
We want:
a) # of teams that include both John and Peter
b) total # of 5person 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 teammembers 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 5person teams
Select 5 teammembers from the 9 players (this can be accomplished in 9C5 ways)
So, the total # of 5person 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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 GMAT Instructor
 Posts: 15041
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You almost had it.gmattesttaker2 wrote:[email protected] wrote:The combination theory approach:sparkle6 wrote:A basketball coach will select the members of a fiveplayer 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?
We want:
a) # of teams that include both John and Peter
b) total # of 5person 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 teammembers 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 5person teams
Select 5 teammembers from the 9 players (this can be accomplished in 9C5 ways)
So, the total # of 5person 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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[email protected] wrote:You almost had it.gmattesttaker2 wrote:[email protected] wrote:The combination theory approach:sparkle6 wrote:A basketball coach will select the members of a fiveplayer 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?
We want:
a) # of teams that include both John and Peter
b) total # of 5person 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 teammembers 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 5person teams
Select 5 teammembers from the 9 players (this can be accomplished in 9C5 ways)
So, the total # of 5person 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
 [email protected]
 GMAT Instructor
 Posts: 15041
 Joined: 08 Dec 2008
 Location: Vancouver, BC
 Thanked: 5254 times
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Glad to helpgmattesttaker2 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
Cheers,
Brent