It is currently 13 Dec 2017, 11:04

Decision(s) Day!:

CHAT Rooms | Ross R1 | Kellogg R1 | Darden R1 | Tepper R1

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

A teacher will pick a group of 4 students from a group of 8

Author Message
TAGS:

Hide Tags

Manager
Joined: 19 Feb 2009
Posts: 54

Kudos [?]: 139 [1], given: 8

A teacher will pick a group of 4 students from a group of 8 [#permalink]

Show Tags

05 Feb 2010, 13:32
1
KUDOS
3
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

72% (01:31) correct 28% (01:39) wrong based on 193 sessions

HideShow timer Statistics

A teacher will pick a group of 4 students from a group of 8 students that includes Bart and Lisa. If one of all the possible four-student groups is picked at random, what is the probability of picking a group that includes both Bart and Lisa?

A) 1/7
B) 3/14
C) 1/4
D) 1/3
E) 3/7
[Reveal] Spoiler: OA

_________________

Working without expecting fruit helps in mastering the art of doing fault-free action !

Kudos [?]: 139 [1], given: 8

Math Expert
Joined: 02 Sep 2009
Posts: 42583

Kudos [?]: 135519 [1], given: 12697

Show Tags

05 Feb 2010, 13:50
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
amod243 wrote:
A teacher will pick a group of 4 students from a group of 8 students that includes Bart and Lisa. If one of all the possible four-student groups is picked at random, what is the probability of picking a group that includes both Bart and Lisa?

A) 1/7
B) 3/14
C) 1/4
D) 1/3
E) 3/7

I am finding this problem very challenging.
dont even have OA for this one.

Can any body please explain me in brief how to approach this one.

$$\frac{C^2_2*C^2_6}{C^4_8}=\frac{3}{14}$$, where $$C^2_2$$ is # of combinations of choosing Bart and Lisa out of Bart and Lisa, which is obviously 1; $$C^2_6$$ # of combinations of choosing 2 other members out of 6 members left; $$C^4_8$$ total # of combinations of choosing 4 people out of 8.

Another approach $$\frac{4!}{2!}*\frac{1}{8}*\frac{1}{7}*\frac{6}{6}*\frac{5}{5}=\frac{3}{14}$$; this is direct calculations of probability 1/8 choosing Bart, 1/7 choosing Lisa, 6/6 and 5/5 any member for group of four; we must multiply this by 4!/2! as this scenario, scenario of $$\{BL**\}$$ can occur in 4!/2! # of ways, which is basically the # of permutations of the symbols $$\{BL**\}$$.
_________________

Kudos [?]: 135519 [1], given: 12697

Senior Manager
Joined: 19 Nov 2007
Posts: 457

Kudos [?]: 225 [0], given: 4

Show Tags

06 Feb 2010, 07:44
amod243 wrote:
A teacher will pick a group of 4 students from a group of 8 students that includes Bart and Lisa. If one of all the possible four-student groups is picked at random, what is the probability of picking a group that includes both Bart and Lisa?

A) 1/7
B) 3/14
C) 1/4
D) 1/3
E) 3/7

I am finding this problem very challenging.
dont even have OA for this one.

Can any body please explain me in brief how to approach this one.

First of all, 4 people can be picked from 8 people in 8C4 ways.
Next, the number of different groups with B and L included in those grps is 6C2.
Hence, required probability is

6C2
___

8C4

Ans is B.

_________________

-Underline your question. It takes only a few seconds!
-Search before you post.

Kudos [?]: 225 [0], given: 4

Manager
Joined: 04 Feb 2010
Posts: 108

Kudos [?]: 10 [0], given: 14

Schools: USC(CGSM), NYU(Int), Texas(CGSM), Ross(CGSM), LBS(Int), Kellogg(Int,JDMBA)

Show Tags

06 Feb 2010, 09:53
I get the sense that the answer is in fact 3/7. Please explain why I would be wrong. I followed the same logic as the poster above me except that for 6C2,there are two ways BL or LB from reducing the pool, so in fact your answer would be 15*2/70 = 3/7

Kudos [?]: 10 [0], given: 14

Senior Manager
Joined: 19 Nov 2007
Posts: 457

Kudos [?]: 225 [0], given: 4

Show Tags

06 Feb 2010, 12:40
JoeMahmah wrote:
I get the sense that the answer is in fact 3/7. Please explain why I would be wrong. I followed the same logic as the poster above me except that for 6C2,there are two ways BL or LB from reducing the pool, so in fact your answer would be 15*2/70 = 3/7

I believe BL and LB are the same since this is a combinations problem.
_________________

-Underline your question. It takes only a few seconds!
-Search before you post.

Kudos [?]: 225 [0], given: 4

SVP
Joined: 08 Jul 2010
Posts: 1857

Kudos [?]: 2400 [2], given: 51

Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: A teacher will pick a group of 4 students from a group of 8 [#permalink]

Show Tags

29 Jun 2015, 21:21
2
KUDOS
Expert's post
amod243 wrote:
A teacher will pick a group of 4 students from a group of 8 students that includes Bart and Lisa. If one of all the possible four-student groups is picked at random, what is the probability of picking a group that includes both Bart and Lisa?

A) 1/7
B) 3/14
C) 1/4
D) 1/3
E) 3/7

Probability = Favorable Outcomes / Total Outcomes

Total Outcomes = Total No. of ways of Picking Group of 4 out of 8 = 8C4 = 8! / (4! * 4!) = 70

Favorable Outcomes = Total No. of ways of Picking Group of 4 out of 8 such that B and L are always in the group (i.e. we only have to pick remaining two out of remaining 6 as B and L must be there is group) = 6C2 = 15

Hence, Probability = 15/70 = 3/14

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Kudos [?]: 2400 [2], given: 51

Intern
Joined: 17 Aug 2016
Posts: 49

Kudos [?]: 3 [0], given: 82

Re: A teacher will pick a group of 4 students from a group of 8 [#permalink]

Show Tags

05 Nov 2016, 05:32
Bunuel wrote:
amod243 wrote:
A teacher will pick a group of 4 students from a group of 8 students that includes Bart and Lisa. If one of all the possible four-student groups is picked at random, what is the probability of picking a group that includes both Bart and Lisa?

A) 1/7
B) 3/14
C) 1/4
D) 1/3
E) 3/7

I am finding this problem very challenging.
dont even have OA for this one.

Can any body please explain me in brief how to approach this one.

$$\frac{C^2_2*C^2_6}{C^4_8}=\frac{3}{14}$$, where $$C^2_2$$ is # of combinations of choosing Bart and Lisa out of Bart and Lisa, which is obviously 1; $$C^2_6$$ # of combinations of choosing 2 other members out of 6 members left; $$C^4_8$$ total # of combinations of choosing 4 people out of 8.

Another approach $$\frac{4!}{2!}*\frac{1}{8}*\frac{1}{7}*\frac{6}{6}*\frac{5}{5}=\frac{3}{14}$$; this is direct calculations of probability 1/8 choosing Bart, 1/7 choosing Lisa, 6/6 and 5/5 any member for group of four; we must multiply this by 4!/2! as this scenario, scenario of $$\{BL**\}$$ can occur in 4!/2! # of ways, which is basically the # of permutations of the symbols $$\{BL**\}$$.

Hi Bunuel, while I understand your approach I was expecting to arrive to the same result with the complementary formula. 1- probability of no having a group with Lisa and Bart.

But I do something wrong. Isn't the probability of a group without Bart and Lisa (6C4)/(8C4)? if I do 1 - (6C4)/(8C4) the final result is wrong...

Kudos [?]: 3 [0], given: 82

Math Expert
Joined: 02 Sep 2009
Posts: 42583

Kudos [?]: 135519 [1], given: 12697

Re: A teacher will pick a group of 4 students from a group of 8 [#permalink]

Show Tags

05 Nov 2016, 06:02
1
KUDOS
Expert's post
bazu wrote:
Bunuel wrote:
amod243 wrote:
A teacher will pick a group of 4 students from a group of 8 students that includes Bart and Lisa. If one of all the possible four-student groups is picked at random, what is the probability of picking a group that includes both Bart and Lisa?

A) 1/7
B) 3/14
C) 1/4
D) 1/3
E) 3/7

I am finding this problem very challenging.
dont even have OA for this one.

Can any body please explain me in brief how to approach this one.

$$\frac{C^2_2*C^2_6}{C^4_8}=\frac{3}{14}$$, where $$C^2_2$$ is # of combinations of choosing Bart and Lisa out of Bart and Lisa, which is obviously 1; $$C^2_6$$ # of combinations of choosing 2 other members out of 6 members left; $$C^4_8$$ total # of combinations of choosing 4 people out of 8.

Another approach $$\frac{4!}{2!}*\frac{1}{8}*\frac{1}{7}*\frac{6}{6}*\frac{5}{5}=\frac{3}{14}$$; this is direct calculations of probability 1/8 choosing Bart, 1/7 choosing Lisa, 6/6 and 5/5 any member for group of four; we must multiply this by 4!/2! as this scenario, scenario of $$\{BL**\}$$ can occur in 4!/2! # of ways, which is basically the # of permutations of the symbols $$\{BL**\}$$.

Hi Bunuel, while I understand your approach I was expecting to arrive to the same result with the complementary formula. 1- probability of no having a group with Lisa and Bart.

But I do something wrong. Isn't the probability of a group without Bart and Lisa (6C4)/(8C4)? if I do 1 - (6C4)/(8C4) the final result is wrong...

Opposite events are groups without Bart and Lisa + groups without Bart but with Lisa + groups without Lisa but with Bart.
_________________

Kudos [?]: 135519 [1], given: 12697

SVP
Joined: 08 Jul 2010
Posts: 1857

Kudos [?]: 2400 [1], given: 51

Location: India
GMAT: INSIGHT
WE: Education (Education)
A teacher will pick a group of 4 students from a group of 8 [#permalink]

Show Tags

05 Nov 2016, 06:09
1
KUDOS
Expert's post
bazu wrote:
Bunuel wrote:
amod243 wrote:
A teacher will pick a group of 4 students from a group of 8 students that includes Bart and Lisa. If one of all the possible four-student groups is picked at random, what is the probability of picking a group that includes both Bart and Lisa?

A) 1/7
B) 3/14
C) 1/4
D) 1/3
E) 3/7

I am finding this problem very challenging.
dont even have OA for this one.

Can any body please explain me in brief how to approach this one.

$$\frac{C^2_2*C^2_6}{C^4_8}=\frac{3}{14}$$, where $$C^2_2$$ is # of combinations of choosing Bart and Lisa out of Bart and Lisa, which is obviously 1; $$C^2_6$$ # of combinations of choosing 2 other members out of 6 members left; $$C^4_8$$ total # of combinations of choosing 4 people out of 8.

Another approach $$\frac{4!}{2!}*\frac{1}{8}*\frac{1}{7}*\frac{6}{6}*\frac{5}{5}=\frac{3}{14}$$; this is direct calculations of probability 1/8 choosing Bart, 1/7 choosing Lisa, 6/6 and 5/5 any member for group of four; we must multiply this by 4!/2! as this scenario, scenario of $$\{BL**\}$$ can occur in 4!/2! # of ways, which is basically the # of permutations of the symbols $$\{BL**\}$$.

Hi Bunuel, while I understand your approach I was expecting to arrive to the same result with the complementary formula. 1- probability of no having a group with Lisa and Bart.

But I do something wrong. Isn't the probability of a group without Bart and Lisa (6C4)/(8C4)? if I do 1 - (6C4)/(8C4) the final result is wrong...

There are 4 cases in total
1) group with Bart and Lisa both = 6C2 = 15
2) group with Bart But not Lisa = 6C3 = 20 (choose 3 out of remaining 6 considering bart chosen and Lisa not to be chosen so kept away))
3) group with Lisa But not Bart = 6C3 = 20
4) group without Bart and Lisa both = 6C4 = 15

The probabilities that you need to subtract from 1 are as follows
2) group with Bart But not Lisa = 6C3 = 20 (choose 3 out of remaining 6 considering bart chosen and Lisa not to be chosen so kept away))
3) group with Lisa But not Bart = 6C3 = 20
4) group without Bart and Lisa both = 6C4 = 15

So you also need to take out the case 2 and 3 to find probability of case (1)

i.e. 1 - (6C4 + 2*6C3)/(8C4) will give you the correct answer where 6C3 are the ways to choose three members with Bart chosen in one case and Lisa chosen already in other.

i.e. 1 - (15+40)/70 = 15/70 = 3/14

I hope this helps!!!
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Kudos [?]: 2400 [1], given: 51

Intern
Joined: 17 Aug 2016
Posts: 49

Kudos [?]: 3 [0], given: 82

Re: A teacher will pick a group of 4 students from a group of 8 [#permalink]

Show Tags

06 Nov 2016, 09:03
Sure, thanks both, what a stupid mistake I made :S

Kudos [?]: 3 [0], given: 82

Non-Human User
Joined: 09 Sep 2013
Posts: 14868

Kudos [?]: 287 [0], given: 0

Re: A teacher will pick a group of 4 students from a group of 8 [#permalink]

Show Tags

24 Nov 2017, 08:13
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 287 [0], given: 0

Re: A teacher will pick a group of 4 students from a group of 8   [#permalink] 24 Nov 2017, 08:13
Display posts from previous: Sort by