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

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

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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

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Re: Probability problem [#permalink]

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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 \(\{BL**\}\).
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Re: Probability problem [#permalink]

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New post 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.

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Re: Probability problem [#permalink]

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New post 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

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Re: Probability problem [#permalink]

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

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New post 29 Jun 2015, 21:21
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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

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

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New post 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...

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

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New post 05 Nov 2016, 06:02
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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.
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 05 Nov 2016, 06:09
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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!!!
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Re: A teacher will pick a group of 4 students from a group of 8 [#permalink]

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

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

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