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

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

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.

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.

In your approach you need to add one more event:

3) 1 person from 1-sibling group and next person from 2-sibling group: 4/7*3/6

p = 20/42 + 12/42 = 16/21

Actually, you have three combinations: 1) 21 2) 11 3) 12.

That is one way, I think it is easier to calculate the probabilities than the combinations.

there is 4/7 chance of selecting a person with one sibling initially with a 5/6 probability the second person will not be a sibling.

There is a 3/7 Chance of selecting a person with 2 siblings initially with 4/6 probability the second person will not be a sibling.

Simply adding these probabilities together gives you the answer.

4/7 * 5/6 + 3/7 * 4/6 = 20/42 + 12/42 = 32/42 = 16/21

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

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.....

As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4).
As there are 3 people with exactly 2 siblings each: we have one triple of siblings (5-6-7).

Let's calculate the probability of opposite event and subtract it from 1. Opposite event would be that chosen 2 individuals are siblings.

# of selections of 2 out of 7 - \(C^2_7=21\);
# of selections of 2 siblings - \(C^2_3+C^2_2+C^2_2=3+1+1=5\), here \(C^2_3\) is the # of ways to choose 2 siblings out of siblings 5-6-7, \({C^2_2}\) is the # of ways to choose 2 siblings out of siblings 1-2, and \(C^2_2\) is the # of ways to choose 2 siblings out of siblings 3-4;

\(P=1-\frac{5}{21}=\frac{16}{21}\).

You can check other approaches in my first post.

Hope it's clear.

Bunuel,

can you explain your #3 approach.

Thanks in advance.

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.

Ways to select the those 4 with their sibling?

4/7 x 1/6 = 4/42

Ways to select those 3 with one of their 2 siblings?

3/7 x 2/6 = 6/42

P = 1 - (4/42 + 6/42) = 1 - 10/42 = (42 -10)/42 = 32/42 = 16/21

P = 16/21

Answer: E

Hi,

Que:
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?

Case 1:
4 people have exactly 1 sibling :
Let A, B, C, D be those 4 people
If A is sibling of B, then B is also a sibling of A.
Basically, they have to exist in pairs
Therefore, these four people compose of two pairs of siblings.

Case 2:
3 people have exactly 2 siblings :
This must obviously be a triplet of siblings.
Let them be E, F, G.
If E is sibling of F, and also is sibling of G., it means F and G are also siblings of each other.
E is sibling of exactly 2 : F and G
F is sibling of exactly 2 : E and G
G is sibling of exactly 2 : E and F
This group has three pairs of siblings.

Now, selecting two individuals out of the group of 7 people has:
1) Two Pairs, as in Case 1.
2) Three Pairs, as in Case 2.
So, 5 cases out of a total of 7C2 = 21 cases.

Hope it clarifies. If you have any further doubt, please point to exactly where you are having a problem understanding it.

It is first important to understand the problem. So let us first assume a specific case.

1. Assume the 7 people are A,B, C, D, E, F and G
2. Assume the 4 people who have 1 sibling are A, B, C and D
3. Let's assume A's sibling is B. Therefore B's sibling is A. Similarly for C and D.
4. So we are left with E, F and G. Each should have exactly 2 siblings.
5. E's siblings will be F and G. So F's siblings will be E and G and G's siblings will be E and F
6. Now look at the general case.
7.Total number of ways of selecting 2 people out of 7 people is 7C2=21
8. Instead of A, B, C and D assume any 4 people. We can see for every such 4 people assumed, there are 2 cases where the selected 2 will be siblings. In the case we assumed they are A and B or C and D. This gives one of the favorable outcomes
9. Or the 2 people selected being siblings may come out of the 3 siblings. The number of favorable outcomes is 3 as we can see in the specific case they are E and F, or F and G or E and G.
10. The total number of favorable outcomes for the selected two being siblings is 2+3=5.
11. The probability that the two selected are siblings is 5/21.
12, Therefore the probability that the two selected are not siblings is 1-5/21= 16/21

Can anyone please help me on where I went wrong?

When you select groups 1 and 2, you have 3*2 choices, when groups 1 and 3 you have 3*2 choices and when groups 2 and 3 you have 2*2 choices: total = 6 + 6 + 4.

First we need to recognize that the given information tells us that the 7 people consist of:
- a sibling trio
- a sibling pair
- and another sibling pair

Let's use counting techniques to answer this question

For this question, it's easier to find the complement.
So P(not siblings) = 1 - P(they are siblings)

P(they are siblings) = [# of ways to select 2 siblings] / [total # of ways to select 2 people]

# of ways to select 2 siblings
Case a) 2 siblings from the sibling trio: from these 3 siblings, we can select 2 siblings in 3C2 ways (3 ways)
Case b) 2 siblings from first sibling pair: we can select 2 siblings in 2C2 ways (1 way)
Case c) 2 siblings from second sibling pair: we can select 2 siblings in 2C2 ways (1 way)

So, total number of ways to select 2 siblings = 3+1+1 = 5

total # of ways to select 2 people
We have 7 people and we want to select 2 of them
We can accomplish this in 7C2 ways (21 ways)

So, P(they are siblings) = 5/21

This means P(not siblings) = 1 - 5/21
= 16/21

Answer: E

Cheers,
Brent

Let A, B, C, D, E, F, and G be the 7 people in the room. To satisfy the condition that 4 people have exactly 1 sibling and 3 people have exactly 2 siblings, we can let A and B be siblings (but not to other people), C and D be siblings (but not to other people), and E, F and G are siblings (but not to other people).

Let’s consider the probability of how each person is chosen:

If A is chosen first, then B can’t be chosen. So the probability is:

1/7 x 5/6 = 5/42

This probability will be the same if B, C, or D is chosen first.

If E is chosen first, then neither F nor G can be chosen. So the probability is:

1/7 x 4/6 = 4/42

This probability will be the same if F or G is chosen first.

Therefore, the overall probability is:

5/42 x 4 + 4/42 x 3 = 20/42 + 12/42 = 32/42 = 16/21

Alternate Solution:

Notice that 2 people can be chosen out of 7 people in 7C2 = 7!/(5!*2!) = (7 x 6)/2 = 21 ways.

With A, B, C, D, E, F, and G as above, we see that there are 5 ways to choose a sibling pair: A-B, C-D, E-F, E-G and F-G. Thus, 21 - 5 = 16 choices of do not include a sibling pair. Therefore, the probability that the chosen two people are not siblings is 16/21.

Answer: E