Siblings Problem - Combinatorics

[day.ash]

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


MGMAT's answer is E - 16/21.

I picked C - 4/7.
My solution was
There are 21 ways to pick 2 people out of 7.

Now out of 7, 4 people are in the one 1 sibling group.
3 people are in the 4 sibling group.

The probability of picking 2 people who are not sibling will be the probability of picking one from each of these two groups.
Number of ways to do so = 3X4= 12

Probability = 12/21 = 4/7.

I just don't get it that is what is missing from my analysis which makes this the wrong answer.

Here is MGMAT's explanation, which I am unable to use to understand mistake in my analysis.


We are told that 4 people have exactly 1 sibling. This would account for 2 sibling relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 siblings. This would account for another 3 sibling relationships (e.g. EF, EG, and FG). Thus, there are 5 total sibling relationships in the group.

Additionally, there are (7 x 6)/2 = 21 different ways to chose two people from the room.

Therefore, the probability that any 2 individuals in the group are siblings is 5/21. The probability that any 2 individuals in the group are NOT siblings = 1 - 5/21 = 16/21.

The correct answer is E.

[rmg]

got the same problem here; i don't understand the mgmat explaination

[hbvankleffens]

Alternative solution:
7 people in a room, exactly 4 people have one sibling: AB & CD, 3 people have two siblings: EFG. What is the chance to randomly select 2 people, which are not siblings?

→ One should select either 1 from AB, 1 from CD, and 0 from EFG, or 1 from
EFG and 1 from ABCD (both divided by all possible combinations 7com2):

( (2com1)(2com1)(3com0)+(3com1)(4com1) ) / (7com2) =(4 +12)/21 = 16/21

[RonPurewal]

your mistake is here:

Now out of 7, 4 people are in the one 1 sibling group.
3 people are in the 4 sibling group.

The probability of picking 2 people who are not sibling will be the probability of picking one from each of these two groups.

this is an incorrect analysis, because you're neglecting the fact that you can also choose two non-siblings out of that first group of four.

remember that each of those four people has only ONE sibling in the room. the only way that's really going to work is like this: (i'm going to use "brother" instead of "sibling", just to make it easier to understand)
A is B's brother
B is A's brother
C is D's brother
D is C's brother
your argument neglects the fact that you could pick, say, A and C (who both come from the group of four), and still satisfy the condition of the problem.

[ps63739]

Hey hbvankleffens .

As you mentioned select 2com1*2com1. I don't understand this..

AB and CD are sibling pairs. So there is only 2 possibilities we can select 1 - AC and second BD. However, your solution mentions 4?

[ps63739]

Hey. Sorry disregard my previous post, I got it AC, AD, BC, and BD.

Sorry about that.

[tim]

The big thing to keep in mind here is that you're not splitting into a group of 3 and a group of 4. 4 people have only one sibling, which gives us two pairs of siblings. So you're splitting your 7 into groups of 3, 2, and 2:

7!/(3!2!2!)

In general this is how you want to think of combinations; it's always about splitting a large group into smaller groups, which is why the numbers on the bottom have to add up to the number on top. Then the biggest hurdle is just figuring out how many are in your small groups..

[kldiaz00]

The big thing to keep in mind here is that you're not splitting into a group of 3 and a group of 4. 4 people have only one sibling, which gives us two pairs of siblings. So you're splitting your 7 into groups of 3, 2, and 2:

7!/(3!2!2!)

In general this is how you want to think of combinations; it's always about splitting a large group into smaller groups, which is why the numbers on the bottom have to add up to the number on top. Then the biggest hurdle is just figuring out how many are in your small groups..

I'm so sorry..I still don't understand this.

7!/(3!2!2!)= 210, no?

How does this come into play? sorry if this is redundant.

[tttidwell]

To make things less complicated and simple...write out the combination/possibilities. Remember probability equals the expected number of outcomes/total number of outcomes. You are looking for the total number of outcomes for the grouping of unrelated siblings and the expected number of outcomes for all siblings. Try it. Should take you under a minute to finish.

[tim]

kldiaz, yes that's 210. It doesn't really come into play in the most efficient solution; i was just addressing the side issue of how many ways you could split up these 7 people into sibling groups. Admittedly, not relevant to this problem because there a lot of easier ways to do this one..

[adiagr]

Two ways to do this problem.

There are three groups:
A: Having 3 persons, each a sibling of other 2
B: having 2 siblings
C having 2 siblings.

Suppose I choose 2 persons and both are siblings. How can I do that?

Either I choose from group A, or B or C (Probabilities will be added)

From A:

Favorable ways: Choosing 2 persons out of available 3 persons: 3 ways

Total Ways: Choosing 2 persons out of available 7 persons
(7!/(2! X 5!)) = 21 ways

so from A: (3/21)

From B:

Favorable ways: Choosing 2 persons out of available 2 persons: 1 way

Total Ways: 21

so from B: (1/21)

Similarly from C: (1/21)

Adding these we get (5/21)

Now Ans will be (1- (above probability)) = (16/21)

Will describe 2nd method in following post.

[adiagr]

Method-2

Those familiar with conditional probability may go for this approach:

There are three groups:
A: Having 3 persons, each a sibling of other 2
B: having 2 siblings
C having 2 siblings.

To choose 2 persons such that they are not siblings.

Choosing first person: Can be chosen without restriction so probability is 1.

Choosing second person:

Choosing 2nd person will depend on the selection of first person

If first man is from Group A, Then the other person cannot be from A

Prob of first person to be from A:

Favorable ways: Choosing 1 person out of available 3 persons: 3 ways

Total Ways: Choosing 1 persons out of available 7 persons
= 7 ways

so Prob of first person to be from A: (3/7)

Now once first person is from A, prob that 2nd person shall not be from A is :

Favorable ways: Choosing 1 persons out of available 4 persons (Gr. B + Gr. C): 4 ways

Total Ways: Choosing 1 persons out of available 6 persons
= 6 ways

Prob (4/6)

so combined probability of this case will be: ((3/7) x (4/6))

If you have followed upto this step:

If first person is from B, prob that 2nd person shall not be from B will similarly be:

(2/7) x (5/6)

If first person is from C, prob that 2nd person shall not be from C will be given by: (same as in case of B)

(2/7) x (5/6)


Total comes out to (16/21)

[RonPurewal]

adiagr -- yes, those look good. nice thorough explanations, too.

[adiagr]

adiagr -- yes, those look good. nice thorough explanations, too.

Thanks Ron.

[mschwrtz]

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