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

4 people have exactly 1 sibling in the room
Translate: #1 is siblings with #2. #3 is siblings with #4. Or you could mix the numbers around, but however way you arrange them, there are TWO pairs of siblings in this group.

3 people have exactly 2 siblings in the room
Translate: #1 is siblings with #2 and #3 – they’re all siblings with each other. So this is ONE triplet of siblings.

So in total, you have one pair, one pair, and then one triplet.

You see the word “NOT” in the question – so you can consider doing the popular (1-prob that they ARE siblings) method.

prob (2 picked people are siblings) = ?

Well, 2 people picked can only siblings if they belong in the same pair or same triplet. We already divided up the people in the the pair, pair, and triplet.

So prob (2 picked people are siblings) = (2 choose 2) + (2 choose 2) + (3 choose 2)
Total # of possibilities = (7 choose 2)

In each case you are choosing 2 people. But your broke down the problem from a large set of 7 people to smaller groups of size 4 and size 3.

So 2nCr2 + 2nCr2 + 3nCr2 = 1 + 1 + 3 = 5 ways to pick siblings

Total # of possibilities = 7nCr2 =

7*6* (5!)
———- = 42/2 = 21 possibilities
2 * (5!)

So # of ways to pick siblings is 5 out of a total of 21.

We want the opposite of this so we do 1 – (5/21) =16/21

Having gone through this, note that of all the answer choices – (D) is the last one I would have picked. Usually, with a question like this, if (5/21) is the correct answer, then whatever 1 – (5/21) is will surely also be an answer choice to trick people who forgot to do the 1 – operation.

Looking at the answer choices, I see that (b) 3/7 and (c) 4/7 add up to 1. So those two can potentially be correct.

(A) 5/21 and (e) 16/21 add up to 1, so they can potentially be correct.

Then (d) just comes out of nowhere as 5/7 by itself.

If I were to guess, I’d probably guess between A, B, C, and E and would eliminate D from guessing.

Just our 2 cents. Hope that helps!