In a room filled with 7 people, 4 people have exactly 1 sibling in the

I found this nice explanation on youtube

https://www.youtube.com/watch?v=Lefh_Ywk3qw

We have three groups: AA, BB, CCC
We can "win" in two ways:
1. We select either A or B and then select someone other than their sibling. \(\frac{4}{7}*\frac{5}{6} = \frac{20}{42}\)
OR
2. We select C and then someone other than their sibling. \(\frac{3}{7}*\frac{4}{6} = \frac{12}{42}\)
\(\frac{20}{42}+\frac{12}{42} = \frac{32}{42} = \frac{16}{21}\)
Answer choice E.

We could also have found the probability of "losing" and subtracted from 1.
1. We select either A or B and then select their sibling. \(\frac{4}{7}*\frac{1}{6} = \frac{4}{42}\)
OR
2. We select C and then their sibling. \(\frac{3}{7}*\frac{2}{6} = \frac{6}{42}\)
\(\frac{4}{42}+\frac{6}{42} = \frac{10}{42} = \frac{5}{21}\)
\(1-\frac{5}{21} = \frac{16}{21}\)
Answer choice E.

Could you explain why the grouping into sibling group is not necessary? I see your simplistic approach but I am not able to replicate the #ways to choose 2 siblings that are NOT related without grouping them.
I understand 2C1*2C1 for # ways to pick NOT-siblings from the 4 person w/ 1 sibling group.
add this to
#ways to get 2C1*3C1 to pick NoT-siblings from the 3person with 2 sibling group.
and multiply that by 2 because 3C1 could pair with one of the other pairs with 1 sibling group.

Not sure how to come up with 16 without grouping them like this..spent hours.. intellectual stimulation study method vs. needing to get the target score are at odds.

I'm not sure how to answer a question like this, except by saying that we can get the answer without thinking about that complication. If you think about picking people one at a time, when you pick the first person, there are just two possibilities:

• 4/7 of the time, you pick a person with one sibling (and we thus have a 5/6 probability the second person we pick is not a sibling of the first)
• 3/7 of the time, you pick a person with two siblings (and we thus have a 4/6 probability the second person we pick is not a sibling of the first)

So we can just split the problem into two cases from the outset, and we don't need to think about how the various people can be related -- since we add when we have cases, the answer is just (4/7)(5/6) + (3/7)(4/6).

This answer is probably among the responses somewhere, however:

4 people have 1 sibling, so that means two sibling pairs.

3 people have two siblings. That can be created by a family of A,B and C. Three pairs can be created from this group.

So there are a total of 5 sibling pairs.

The number of ways 2 people can be selected from 7 is:

7!/2!5!= 21
So the probability of selecting a sibling pair is:

5/21 and the probability of drawing a non sibling pair is:
1-5/21 = 16/21

Posted from my mobile device

Hi

Why is the following approach wrong?

Ways to select 1 out of 4 (who have 1 sibling each): 4C1 = 4
Ways to select 1 out of 3 (who have 2 siblings each): 3C1 =3

Desirable outcomes = 4x3 = 12

Total possible outcomes = 7C1 = 21.

So, probability = 12/21 = 4/7

There are several issues with your approach and I recommend studying the combinatorics topics.

First, you would benefit by drawing a diagram of the people and pairs from the wording of the question:

Pair A1,A2
Pair B1, B2
Pairs C1,C2; C1,C3; C2,C3


Your approach starts down the path of finding the number of sibling pairs, which is fine, but you continue this approach through to the answer, but the question is asking for probability of NON sibling pairs. So, you're losing focus.

Back to your starting point.

Your approach for finding the number of sibling pairs among the 4 people is incorrect and is actually the number of ways to select 1 person from 4.

So it counts person A1 and Person A2 as two choices, even though they both could comprise 1 pair, so the approach double counts and the correct number of sibling pairs among the four people is:

2


Your approach for the number of pairs from the 3 people is correct because there ARE 3 pairs and the answer is:

3

Your approach following this is to multiply these results together, which is incorrect.

Multiplying means pairing each answer from one calculation with each answer from the other, which doesn't make sense since it would be forming non-sibling pairs, right ? A1 and C2 would be joined up, but they're not siblings.

So ADDING the results above is the correct approach, equal to:

5

Since these are pairs of people, the probability of these needs to be computed in comparison to the total ways TWO people can be drawn from the 7, or:

7!/2!5! = 21

So the probability of drawing a sibling pair is then:

5/21

Now, back to your final calculation. You stopped at the point above.

Since the question is asking for the probability of NOT drawing a sibling pair, the above needs to be subtracted from 1:

1-5/16 = 11/16

Posted from my mobile device

Hi IanStewart,

(4/7)(5/6) + (3/7)(4/6) = 16/21

In above solution, when we select any person from a group of 4 i.e., 4/7 and next pick any other person who is not the already selected person's sibling we have 5 choices i.e., 5/6. This is clear to me. But where I get a doubt is in the next part [ (3/7)(4/6)]. Here, wouldn't it also include a case where 1 person selected from a group of 3 and next 1 person selected from a group of 4 - that was already considered in 1st portion of the solution?

I think my doubt is that how the order in which the people are selected would matter?

My approach was like this:
(4/7)(2/6) [selecting any one out of 4 grp people and then leaving their sibling, selecting from remaining 2 out of the 4 grp people]
+
(4/7)(3/6) [selecting any one out of 4 grp people and then selecting 1 from 3 grp people]

So there are total 7 persons in the room (lets call them A,B,C,D,E,F,G). 4 persons have EXACTLY 1 sibling and 3 have EXACTLY 2 siblings
Lets say A and B (Group 1) , C and D (Group 2) are 4 persons with EXACTLY 1 sibling. and E,F,G (Group 3) are three siblings, each one of which have EXACTLY 2 siblings
Now to select the group of two , both of which should not be sibling, we need to select the person from different groups.
Group 1 and Group 2 = 2C1 x 2C1 = 4
or
Group 1 and Group 3 = 2C1 x 3C1 = 6
or
Group 2 and Group 3 = 2C1 x 3C1 = 6

Total no of ways = 16

Total no of ways to select two persons from group of 7 = 7C2 = 21
Total prob. = 16/21