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?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

You have people 1-2-3-4-5-6-7

4 have exactly 1 sibling which can mean:
1-2 are siblings
3-4 are siblings

3 have exactly 2 siblings which can mean:
5-6-7 are siblings

Let's start with the group of 4:
The probability of picking 1 is (1/7)
The probability of not getting a sibling pair is (5/6) because the only other sibling is 2
Therefore, if 1 is selected first the probability of not getting a sibling pair is 5/42
Multiply that by 4 because the probability is the same whether you start with 1-2-3-4 so you get 20/42 for the first 4 people

Now let's go to the group of 3:
The probability of picking 5 is (1/7)
The probability of not getting a sibling pair is (4/6) which the non-sibling pair is 1-2-3-4
Therefore the probability is 4/42. Multiply that probability by 3, which represent 5-6-7 so the probability is 12/42

Now you have two probabilities: 12/42 and 20/42
add both and you get 32/42 or 16/21

I also get the same result 16/21, but my approach is a bit different.
There are 7C2 =21 ways to select 2 people in a group of 7.
There are 1+1+3 = 5 ways to select 2 people who are siblings.
So the probability of selecting 2 people who are not siblings is : 1-5/1 =16/21

@ Walker: I also tried to solve the problem in the way that you suggested, but i can't get myself clear why the two events mentioned are mutually exclusive.
I thought the following events are mutually exclusive for this problem, and i arrived with the following solution
1) 1 person from 2-sibling group and 1 person from 1-sibling group, which is the same with your second event:
we have 3/7 probability of getting 2-siblings person and 4/6 probability of not getting his or her sibling.
2) No one from the 2-sibling group.
we have 4/7 probability of getting 1-sibling person and 2/6 probability of not getting his or her sibling and not within one of 2-sibling group.
p = 3/7*4/6 + 4/7*2/6 = 20/42 (this is not within the answers, so i guess it is wrong, but could you please help me where I got wrong with this). Many thanks.

Four people have exactly one sibling and three people have exactly two siblings.

Think about how this can be done...the only way to organize is this is to have two pairs of siblings and one trio of siblings. In other words:

AB (one sibling pair)
CD (another sibling pair)
EFG (a trio of siblings)

This way each of A and B are siblings, each of C and D are siblings, so we have four people each with exactly one sibling. And each of E, F, and G are siblings so we have three people each of whom has exactly two siblings.

Because this is a "NOT" probability question, we should see how many ways we CAN select two individuals who are siblings.

Well, we can select the AB pair or the CD pair. So far, that's 2 ways. We can also select any two people from the EFG trio, so that's another 3C2 or 3 ways.

So, there are a total of 2+3 = 5 ways of pulling out 2 individuals who are siblings.

Probability = (#desired)/(#total)

The denominator of the formula is just all the ways we can select any two people from the seven. So we have:

Probability of selecting 2 people who are siblings = 5/7C2 = 5/21

Therefore, the probability of selecting 2 people who are NOT siblings is:

1 - 5/21 = 16/21

Choose E.

Out of 7 people
4 have 1 sibling So these 4 form 2 pairs of siblings
3 have 2 siblings So all these 3 are siblings to each other.

So the 7 people are like A1A2, B1B2, C1C2C3 where A1 and A2 are siblings to each other , B1 and B2 are siblings to each other and C1,C2 and C3 are siblings to each other.

Probability of selecting 2 people who are not siblings
= 1 - Probability of selecting 2 people who are siblings
= 1 - \((C^2_1+C^3_2)/C^7_2\) (Select either A1A2/B1B2 or any 2 out of C1C2C3)
= 1- 5/21 = 16/21

total possibilities 2C7=21
possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5
probability 1-5/21

vittar..can you please explain 2C3.....

u have 2 groups
(a b x y) and (c d e)


note, that a=b (siblings)
c=d=e (siblings)

to find no two identical siblings, u need to find 2 no siblings in group 1 , or 1 sibling from group 1 and 1 sibling from group 2

1.to find 2 no siblings in group 1 (a b x y)-

u have 4 such combinations - (a;x) (a;y) (b;x) (b;y)

2. 1 sibling from group 1 and 1 sibling from group 2 -

4C1*3C1=4*3=12

(4+12)/7C2=16/21

hope it helps,but if u have any question, please, ask.
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Sure.

We have the following siblings: {1, 2}, {3, 4} and {5, 6, 7}.

Now, in order to select two individuals who are NOT siblings we must select EITHER one from {5, 6, 7} and ANY from {1, 2} or {3, 4} OR one from {1, 2} and another from {3, 4}.

\(P=2*\frac{3}{7}*\frac{4}{6}+2*\frac{2}{7}*\frac{2}{6}=\frac{4}{7}+\frac{4}{21}=\frac{16}{21}\).

3/7 - selecting a sibling from {5, 6, 7}, 4/6 - selecting any from {1, 2} or {3, 4}. Multiplying by 2 since this selection can be don in two ways: the first from {5, 6, 7} and the second from {1, 2} or {3, 4} OR the first from {1, 2} or {3, 4} and the second from {5, 6, 7};

2/7 - selecting a sibling from {1, 2}, 2/6 - selecting a sibling from {3, 4}. Multiplying by 2 since this selection can be don in two ways: the first from {1, 2} and the second from {3, 4} OR the first from {3, 4} and the second from {1, 2}.

Other approaches here: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html#p645861

Hope it's clear.
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I don't understand the red part at all.

As for your solution it's wrong because when for case 1 you select one from {A,B} and then one from {C,D, E, F, G} you can have pair {A,C}. Next, when for case 2 you select one from {C,D} and then one from {A,B,E,F,G} you can have pair {A,C} again. So, you double-count some cases.

Hope it's clear.
_________________

well, this is how i approached it...

Total 7 people - 1,2,3,4,5,6,7
4 people have 1 sibling - [1,2];[3,4] - 2 single-sibling groups
3 people have 2 siblings - [5,6,7] - 1 two-siblings group

Total ways to select 2 people, 7C2 = 21.
Ways to select only siblings : 2C1( ways to select 1 group from 2 single-sibling groups) + 3C2( ways to select 2 people from 2 sibling-group) = 2+3= 5

Probability that NO siblings selected = 1- 5/21 = 16/21.

Hence, D.

Please let me know if my approach is flawed.