Probability

[RobertoB400]

I can't find the mathematical reasoning behind this question despite getting it right both on the CAT and the post-CAT revision. I reckon the answer choices are so logically easy to be crossed out leaving you with the only reasonable answer. But anyways, I would like to fully understand the breakdown process behind this question.

"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?"

5/21
3/7
4/7
5/7
16/21

I grouped the 7 people under 3 groups. 2 groups have a single relationship with 2 people each, and the last group has 3 people with a double relationship each.

Now, how to calculate the ways we can select 2 random people?

And how is that affected by having groups with different amounts of relationships?

Thanks.

[RonPurewal]

I can't find the mathematical reasoning behind this question despite getting it right both on the CAT and the post-CAT revision. I reckon the answer choices are so logically easy to be crossed out leaving you with the only reasonable answer.

^^ i'm curious. can you explain the red thing?

if your implication here is "the probability should be high", then, sure, i'm with you there.
however... choice D is 15/21, and choice E (the correct response) is 16/21. i'm not seeing how "logic" / common sense can differentiate between two choices that are only 1/21 apart.

why is choice D "so logically easy to cross out"?

[RonPurewal]

I grouped the 7 people under 3 groups. 2 groups have a single relationship with 2 people each, and the last group has 3 people with a double relationship each.

this ^^ is the "money step". there's only one configuration that satisfies the requirements--namely, this one.

once you have that, you can just list out the ways of picking 2 people.
let's say we have brothers A and B, sisters C and D, and brothers E, F, and G (it doesn't matter whether they are brothers or sisters).

here are all the ways of picking pairs:
AB AC AD AE AF AG
BC BD BE BF BG
CD CE CF CG
DE DF DG
EF EG
FG

there are 21 pairs, of which only 5 (AB, CD, EF, EG, FG) are brothers/sisters. so, the other 16 out of 21 are not brothers/sisters.

[RonPurewal]

by the way, you should be able to make that list in at most 30 seconds.

if you can't, then you should practice making these kinds of lists.
boring? well, yes.
helpful? absolutely.
with the exception of very basic calculations (e.g., there are 26^4 possibilities for a four-letter code), it's nearly always possible to solve GMAC's combination problems by just listing out the possibilities.

take a look through some GMAC combination problems sometime. you'll notice that the numbers of possibilities are usually ...
... small enough to list and count,
BUT
... large enough to dissuade people who are easily dissuaded.

by contrast, lots of third-party combination problems--including a few of our own, unfortunately--have too many possibilities for counting to be feasible.
we're working on fixing that situation. if you encounter such a problem, don't let it dissuade you from making lists in general.