John and Peter

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ST: Master | Next Rank: 500 Posts; Posts: 191; Joined: Sat Feb 14, 2009 11:56 am; Thanked: 1 times

John and Peter

by ST » Wed Jul 08, 2009 12:41 pm

A basketball coach will select the members of a five-players team from armog 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

Answer is D

Source: Beat The GMAT — Problem Solving | Wed Jul 08, 2009 12:41 pm

dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

Re: John and Peter

by dtweah » Wed Jul 08, 2009 1:13 pm

ST wrote:A basketball coach will select the members of a five-players team from armog 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

Answer is D

(2C2 x 7C3)/9C5=5/18

shibal: Master | Next Rank: 500 Posts; Posts: 345; Joined: Wed Mar 18, 2009 6:53 pm; Location: Sao Paulo-Brazil; Thanked: 12 times; GMAT Score:660

by shibal » Wed Jul 08, 2009 4:37 pm

dtweah, why did you do 2c2*7c3? i disconsidered it and got 5/18. maybe i was lucky but what is the concept behind it?

ST: Master | Next Rank: 500 Posts; Posts: 191; Joined: Sat Feb 14, 2009 11:56 am; Thanked: 1 times

by ST » Wed Jul 08, 2009 4:38 pm

could you please explain it more? How did you get 2C2 and 7C3

imhimanshu: Senior | Next Rank: 100 Posts; Posts: 87; Joined: Sat Feb 28, 2009 7:01 am; Location: India; Thanked: 2 times

by imhimanshu » Wed Jul 08, 2009 7:42 pm

let me explain this:
total number of outcomes=9C5 i.e selecting five persons from 9 guys
number of favourable = 7C3 since jhn and peter are already included. we need to select 3 guys out of remaing 7. i.e 7C3
so probability= 7C3/9C5 =5/18 hence D

dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

by dtweah » Thu Jul 09, 2009 6:06 am

ST wrote:could you please explain it more? How did you get 2C2 and 7C3

Sorry for delay.

Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G

Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5

2/2=2C2=1 x 7C3/9C5

That is the concept behind it.

real2008: Legendary Member; Posts: 527; Joined: Thu May 01, 2008 12:06 am; Thanked: 7 times

by real2008 » Thu Jul 09, 2009 12:29 pm

dtweah wrote:
ST wrote:could you please explain it more? How did you get 2C2 and 7C3
Sorry for delay.

Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G

Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5

2/2=2C2=1 x 7C3/9C5

That is the concept behind it.

why 2C2?

dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

by dtweah » Thu Jul 09, 2009 1:11 pm

real2008 wrote:
dtweah wrote:
ST wrote:could you please explain it more? How did you get 2C2 and 7C3
Sorry for delay.

Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G

Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5

2/2=2C2=1 x 7C3/9C5

That is the concept behind it.
why 2C2?

The number of ways of picking two persons from among 2 persons. I said separate two groups. One group of 2 persons and another group of 7 persons. If the 2 persons must be included then you find that number of ways of selecting them is:
2C2 x number of ways of selecting 3 people from 7=7C3/9C5

skprocks: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Sun Dec 27, 2009 8:50 am; Thanked: 3 times; GMAT Score:710

by skprocks » Sat Jul 24, 2010 11:15 pm

dtweah wrote:
real2008 wrote:
dtweah wrote:
ST wrote:could you please explain it more? How did you get 2C2 and 7C3
Sorry for delay.

Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G

Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5

2/2=2C2=1 x 7C3/9C5

That is the concept behind it.
why 2C2?
The number of ways of picking two persons from among 2 persons. I said separate two groups. One group of 2 persons and another group of 7 persons. If the 2 persons must be included then you find that number of ways of selecting them is:
2C2 x number of ways of selecting 3 people from 7=7C3/9C5

That approach is fine.But when I consider the approach of 1- Negation of prob , I am not getting the expected answer.

P(John and Peter were not selected)= 7c5/9c5
And hence P(John and Peter Selected)=1-7c5/9c5=1=1-1/6=5/6 which is not any of the option.I am unable to understand.
Why this approach does not yield an answer?

skprocks: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Sun Dec 27, 2009 8:50 am; Thanked: 3 times; GMAT Score:710

by skprocks » Sat Jul 24, 2010 11:17 pm

dtweah wrote:
real2008 wrote:
dtweah wrote:
ST wrote:could you please explain it more? How did you get 2C2 and 7C3
Sorry for delay.

Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G

Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5

2/2=2C2=1 x 7C3/9C5

That is the concept behind it.
why 2C2?
The number of ways of picking two persons from among 2 persons. I said separate two groups. One group of 2 persons and another group of 7 persons. If the 2 persons must be included then you find that number of ways of selecting them is:
2C2 x number of ways of selecting 3 people from 7=7C3/9C5

outreach: Legendary Member; Posts: 748; Joined: Sun Jan 31, 2010 7:54 amFollowed by: 3 members; Thanked: 46 times

by outreach » Sun Jul 25, 2010 1:49 am

opposite event will be: none of them is in the team + only one of them is in the team

skprocks wrote:
dtweah wrote:
real2008 wrote:
dtweah wrote:
ST wrote:could you please explain it more? How did you get 2C2 and 7C3
Sorry for delay.

Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G

Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5

2/2=2C2=1 x 7C3/9C5

That is the concept behind it.
why 2C2?
The number of ways of picking two persons from among 2 persons. I said separate two groups. One group of 2 persons and another group of 7 persons. If the 2 persons must be included then you find that number of ways of selecting them is:
2C2 x number of ways of selecting 3 people from 7=7C3/9C5
That approach is fine.But when I consider the approach of 1- Negation of prob , I am not getting the expected answer.

P(John and Peter were not selected)= 7c5/9c5
And hence P(John and Peter Selected)=1-7c5/9c5=1=1-1/6=5/6 which is not any of the option.I am unable to understand.
Why this approach does not yield an answer?

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by GMATGuruNY » Sun Jul 25, 2010 2:05 am

skprocks wrote:
dtweah wrote:
real2008 wrote:
dtweah wrote:
ST wrote:could you please explain it more? How did you get 2C2 and 7C3
Sorry for delay.

Whenever you told that "JOhn and James must be included" you can treat that as a Sure EVENT. It has to happen. The probably of a Sure event is 1. Separate the sure event from the others. You have J J A B C D E F G

Probability of choosing james and john x probability choosing the other 3 after you have chosen james and John / 9C5

2/2=2C2=1 x 7C3/9C5

That is the concept behind it.
why 2C2?
The number of ways of picking two persons from among 2 persons. I said separate two groups. One group of 2 persons and another group of 7 persons. If the 2 persons must be included then you find that number of ways of selecting them is:
2C2 x number of ways of selecting 3 people from 7=7C3/9C5
That approach is fine.But when I consider the approach of 1- Negation of prob , I am not getting the expected answer.

P(John and Peter were not selected)= 7c5/9c5
And hence P(John and Peter Selected)=1-7c5/9c5=1=1-1/6=5/6 which is not any of the option.I am unable to understand.
Why this approach does not yield an answer?

The approach you're using is:

P(good outcome) = 1 - P(bad outcome)

To use this approach, we have to account for ALL the different ways to get a bad outcome:

P(not both J and P) = P(not J and not P) + P(J or P and 4 others)

P(not J and not P) = 7c5/9c5 = 1/6
P(J or P and 4 others) = 2 * 7c4/9c5 = 5/9
P(not both J and P) = 1/6 + 5/9 = 13/18

So P(both J and P) = 1 - 13/18 = 5/18.

I'd use this approach only when P(bad outcome) is easier to determine than P(good outcome). In this problem, P(good outcome) involves less math.

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