In a room filled with 7 people, 4 people have exactly 1 sibl

[lina.fan]

Hi,

Can someone help better explain the answer to this question? I was not able to understand the Manhattan GMAT CAT explanation.

Thanks,
Lina

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?
5/21
3/7
4/7
5/7
16/21

[LazyNK]

Hey Lina,
To understand better, lets count.

4 people have one sibling.
So Lets consider one person of these 4, say A.
He has one sibling, say B, which is also a part of these 4 people.
Similarly, there would be a third person say C in this group, who'd have another person say D, also a part of these 4, as is siling.
So we created following sibling pairs :
AB
CD
and that completes 4 people who have one sibling only.

Also, 3 people have 2 siblings each.
Lets take one person E of this group. As he has 2 siblings, he'd have another two people say F & G, a part of this group of 3 people, as his siblings.
So we create the following sibling triplet :
EFG
and that completes the three people who have two siblings each, and also completes the group of 7 people viz. AB, CD and EFG.

Total ways of choosing 2 people at random from 7 people = 7C2 = 7.6/2=21 ---1
Total ways of choosing 2 people such that they are siblings is = (no. of ways of choosing 2 individuals from pair AB) + (no. of ways of choosing 2 individuals from pair CD) + (no. of ways of choosing 1 individual from EFG = 1 + 1 + 3C2=5 ---2
So number of ways of choosing 2 people such that they are not siblings= equation(1)-equation(2)=21-5=16
So probability of 2 people not being siblings=16/21

-NK

[tim]

thanks Lazy..

Lina, let us know if you have further questions on this one. If so, please be more specific about what you didn't understand rather than just asserting that you didn't get our explanation..

[krishnan.anju1987]

Hi,

I did this problem using the above mentioned approach and also the usual probability approach using the formula

P(choosing a person with 1 sibling)*P(choosing someone apart from his sibling)+P(choosing someone who has two siblings)*P(choosing someone apart from his siblings)

However, when I tried the below mentioned approach, I got stuck.

I tried using the combination formula

7C1 * 5C1 + 7C1*5C1 + 7C1*4C1/7C2

I am sure I am missing out something that I am supposed to divide these numbers with but try as I might, I am unable to figure it out.

Any help would be appreciated.

[tim]

for one thing, all your 7C1's are incorrect; where did they come from? also, your mathematical expression has more terms than your English sentence, so you may want to check your work..

[jamiefurbush]

Total ways of choosing 2 people at random from 7 people = 7C2 = 7.6/2=21 ---1

-NK

Can you please explain why you divide by 2 in this equation?

Best,
Jamie

[tim]

if we are choosing 2 people from a group of 7, we need to split the group of 7 into 5 that we choose and 2 that we don't choose. this means we put 7! on the top of the fraction and 5! and 2! on the bottom. as the previous post indicates, the 7!/5! reduces to 7*6, but we still have the 2! to contend with. this is where dividing by 2 comes in..

[vikash.121186]

Hi,

I did this problem using the above mentioned approach and also the usual probability approach using the formula

P(choosing a person with 1 sibling)*P(choosing someone apart from his sibling)+P(choosing someone who has two siblings)*P(choosing someone apart from his siblings)

However, when I tried the below mentioned approach, I got stuck.

I tried using the combination formula

7C1 * 5C1 + 7C1*5C1 + 7C1*4C1/7C2

I am sure I am missing out something that I am supposed to divide these numbers with but try as I might, I am unable to figure it out.

Any help would be appreciated.

The numerator part of the solution will count the same solution (favourable cases) multiple number of times and hence there will be over counting.

For ex: selecting A and then B is same as selecting B and then A, if the questions asks about selecting 2 people from a lot of 10 different guys.

[tim]

thanks!