Math Question - Seeking the Wizard/ or anyone that can better explain this to me

I came across a question in my studies and I'm a bit confused. Perhaps someone can better explain it to me.

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?

1.) What the heck does this question mean? I'm not great with family relationships, but I'm not sure where to begin breaking down this question other the 7C2 = 21 ways to choose 2 ppl from a set 7.

2.) I'm told that the answer is 16/21. Based on the fact that 4p to 1s = 2 relations and 3p to 2s = 3 relations. Therefore 5/21 = ppl w/ sibling relations and 1 - 5/21 = 16/21. Now, what are they talking about here?

3.) Would you consider this to be a poorly worded question? I'm told its an advanced level question, but I can't even figure out what the question is suppose to mean.

Would really appreciate your help on this. Can someone translate this into more laymen's terms. I'm feeling a bit dumbfounded.

I'll admit it's a bit of a confusing question.... After some consideration, I think there's only one way the setup can be interpreted:

2 people form a sibling pair, that is, they share Mom A and Dad A.
2 people form an additional sibling pair, sharing Mom B and Dad B
3 people form a sibling triplet, all sharing Mom C and Dad C.

With that as the setup, I'll admit to being too lazy to go through the detailed computations to get to the final answer, but I think it can be calculated.

Correct me if I'm wrong, but... it may be even simpler than I first suspected...

3/7(chance that first person is from triplet)*2/6(chance that 2nd person is from triplet) + 4/7(chance that first person is from a pair)*1/6(chance that second person is from same pair)

6/42 + 4/42 = 10/42 = 5/21

And re-read the question to make sure you understand it...

Chance of NOT siblings = 1-chance of siblings (calculated above) = 16/21

Quote: CapnDave

I'll admit it's a bit of a confusing question.... After some consideration, I think there's only one way the setup can be interpreted:

2 people form a sibling pair, that is, they share Mom A and Dad A.
2 people form an additional sibling pair, sharing Mom B and Dad B
3 people form a sibling triplet, all sharing Mom C and Dad C.

With that as the setup, I'll admit to being too lazy to go through the detailed computations to get to the final answer, but I think it can be calculated.

As previously stated, there are 21 combinations of two from the pool of 7. However, there are only five combinations that produce siblings:
A1/A2 = sibling pair A
B1/B2 = sibling pair B
C1/C2 = a sibling pair from the three C's
C2/C3 = a sibling pair from the three C's
C1/C3 = a sibling pair from the three C's

Therefore there are 16 ways to pull two that do not contain siblings.

Hope that helps.

I think the wording is fine, and I agree that the answer is 16/21. There are combin(7,2)=21 ways to pick 2 people out of the 7. Think of the sibling sets as AABBCCC.

There are 2*2=4 ways to choose an A and B.
There are 2*3=6 ways to choose an A and C.
There are 2*3=6 ways to choose a B and C.

So the probability is (4+6+6)/combin(7,2) = 16/21.

The unspoken challenge in this question is deducing the relationships between the seven individuals. Once you figure that out, the rest falls into place.

Thank you all for replying. Now my head is really spinning. I'm going to have to reread this a couple of times over. I'm still not seeing it as far as fully understanding it.

Oh and one more thing, I was told that this question s/b done within 2 minutes. Did you all get this within that time period? I spent considerably more trying to get a clue as to what the question was talking about......

The wording is really getting to me. When they say 4 ppl have 1 sibling in the room, does that mean both mom or dad has their child in the room. Would you then count mom or dad + child as ppl in the room?

3ppl have 2 siblings in the room. Who is actually in the room? The parents or the siblings or both.....?

No parents around. Two sets of twins, one set of triplets. These sets of siblings don't really have to have the same birthday. And each set is from a different family.

I think you might be getting tripped up on the definition of "Sibling". It means children from common parents. I.E. Brothers and/or sisters.

The first important clue is the second clue provided: THREE people that have TWO siblings. Ding! That means that three people are from the same parents.

That leaves four people. And the clue that four have one sibling. That means two pairs of brothers/sisters.

Once you figure out this relatively simple problem, then expanding it for a larger group is easy.

If a clue is that 'x' people have 2 siblings, then 'x' better be a multiple of three...

Ok that's almost starting to make sense to me. To clarify better.....would it be better to say that Three people ARE siblings instead of Three people have 2 siblings. I'm guessing that is suppose to mean the same thing right?

I think I'm getting it.....

Quote: Asswhoopermcdaddy

Ok that's almost starting to make sense to me. To clarify better.....would it be better to say that Three people ARE siblings instead of Three people have 2 siblings. I'm guessing that is suppose to mean the same thing right?

I think I'm getting it.....

That is the logical part of the problem. If you have three people that are siblings it is logically equivalent to three people have 2 siblings if and only if they are not a sibling of the other four people in the room.

The other 4 people have to be two pairs of siblings, so that these 4 people have exactly 1 sibling in the room.

To be a pure mathematician you should prove that no other combination of siblings also meets the criteria, but you can probably ignore that step.

Now you have
three children from the set of parents designated A
two children from the set of parents designated B
two children from the set of parents designated C

The Wizard's combinatorial calculation let's you answer the probability question.