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A teacher will pick a group of 4 students from a group of 8 05 Feb 2010, 14:32
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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
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A teacher will pick a group of 4 students from a group of 8 05 Feb 2010, 14:50
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amod243
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.
Show more

\(\frac{C^2_2*C^2_6}{C^4_8}=\frac{3}{14}\), 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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A teacher will pick a group of 4 students from a group of 8 06 Feb 2010, 08:44
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amod243
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.
Show more

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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A teacher will pick a group of 4 students from a group of 8 06 Feb 2010, 10:53
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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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A teacher will pick a group of 4 students from a group of 8 06 Feb 2010, 13:40
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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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I believe BL and LB are the same since this is a combinations problem.
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A teacher will pick a group of 4 students from a group of 8 29 Jun 2015, 22:21
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amod243
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
Show more

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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A teacher will pick a group of 4 students from a group of 8 05 Nov 2016, 06:32
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amod243
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 \(\{BL**\}\).
Show more


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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A teacher will pick a group of 4 students from a group of 8 05 Nov 2016, 07:02
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amod243
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...
Show more

Opposite events are groups without Bart and Lisa + groups without Bart but with Lisa + groups without Lisa but with Bart.
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A teacher will pick a group of 4 students from a group of 8 05 Nov 2016, 07:09
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bazu
Bunuel
amod243
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...
Show more

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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A teacher will pick a group of 4 students from a group of 8 06 Nov 2016, 10:03
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Sure, thanks both, what a stupid mistake I made :S
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A teacher will pick a group of 4 students from a group of 8 02 Sep 2020, 22:56
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GMATinsight
amod243
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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Please explain this 6C2 concept

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A teacher will pick a group of 4 students from a group of 8 12 Oct 2020, 07:43
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Probability = favorable outcomes / total number of outcomes.

8C4 = 70 total outcomes.

Since Bart & Lisa MUST be included, we leave the two out. We have 6 remaining; the number of ways to choose 2 people from the 6 remaining are: 6C2 = 15

15/70 = 3/14
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A teacher will pick a group of 4 students from a group of 8 12 Oct 2020, 07:52
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Another approach:

Probability of choosing Bart first: 1/8

Probability of choosing Lisa second: 2/7

Probability of choosing a third person: 6/6

Probability of choosing a 4th person: 5/5

(1/8) * (2/7) * (6/6) * (5/5) = 1/28

Now since these events are mutually exclusive, the order does not matter. Therefore we need to multiply 1/28 by 4C2 = 6/28 = 3/14
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A teacher will pick a group of 4 students from a group of 8 20 Feb 2021, 14:38
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Bunuel
amod243
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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Why are we not multiplying numerator twice since the two fixed members can be selected either ways?
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A teacher will pick a group of 4 students from a group of 8 26 Feb 2021, 21:48
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Bart and lisa getting picked is fixed, hence effectively all you are choosing is 2 other people out of the 6 left out people.

Total ways of choosing 4 out of 8 people = 8C4

Therefore, 6C2 / 8C4 = 3/14
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A teacher will pick a group of 4 students from a group of 8 26 Dec 2023, 01:33
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