Tricky probability

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sparkle6: Senior | Next Rank: 100 Posts; Posts: 54; Joined: Tue Sep 20, 2011 7:58 am; Thanked: 1 times; Followed by:1 members

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]

Quote

Source: Beat The GMAT — Problem Solving | Thu Sep 29, 2011 6:52 am

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » 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

Brent Hanneson - Creator of GMATPrepNow.com

Quote

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » 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

Brent Hanneson - Creator of GMATPrepNow.com

Quote

gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pm; Thanked: 11 times; Followed by:8 members

by gmattesttaker2 » Sun Jul 08, 2012 11:45 pm

Brent@GMATPrepNow 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

Quote

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Mon Jul 09, 2012 6:28 am

gmattesttaker2 wrote:
Brent@GMATPrepNow 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

You almost had it.

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

Brent Hanneson - Creator of GMATPrepNow.com

Quote

gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pm; Thanked: 11 times; Followed by:8 members

by gmattesttaker2 » Mon Jul 09, 2012 11:13 pm

Brent@GMATPrepNow wrote:
gmattesttaker2 wrote:
Brent@GMATPrepNow 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
You almost had it.

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

Quote

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » 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

Glad to help