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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