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

Question

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

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

Bunuel · Accepted Answer

\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**\} .

GMATinsight · Answer

Probability = Favorable Outcomes / Total Outcomes

Total Outcomes  = Total No. of ways of Picking Group of 4 out of 8 = 8 C 4 = 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) = 6 C 2 = 15

Hence,  Probability  =  15 / 70  =  3/14

Answer: Option B

vscid · Answer

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.

JoeMahmah · Answer

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

vscid · Answer

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

bazu · Answer

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...

Bunuel · Answer

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

GMATinsight · Answer

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!!!

bazu · Answer

Sure, thanks both, what a stupid mistake I made :S

shreyasaxena2301 · Answer

. 
Please explain this 6C2 concept

Posted from my mobile device

Basshead · Answer

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

Basshead · Answer

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

lakshya14 · Answer

Why are we not multiplying numerator twice since the two fixed members can be selected either ways?

wishmasterdj · Answer

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

pappal · Answer

total no. of groups in which 4 students are selected out of 8 at random =8C4
now no. of groups in which B & C are there means no. of ways of selecting 2 students from remaining 06 students=6C2
so required probability=6C2/8C4=3/14

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards