BTGmoderatorLU wrote: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

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

Using counting techniques:

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

Hi All,

Brent's approach focuses on the use of the Combination Formula, which is probably the most "elegant" way to solve this problem. There are other ways to solve this problem though, if you're not yet comfortable with doing that type of math. Here's an option that focuses on straight "probability math"....

The first step is to organize the 7 people into family-groups. 4 people have exactly 1 sibling. We'll call those people:

A1, A2
B1, B2

A1 and A2 are siblings, etc.

3 people have exactly 2 siblings. We'll call them:

C1, C2 and C3

We're going to select 2 people from the room; what is the probability that they are NOT siblings.

Probability = (# of ways you want something to happen)/(# of total possibilities)

The total number of possibilities = (7)(6) = 42

Now let's figure out the ways that give us 2 non-siblings:

If the first person selected is:
A1, then there are 5 non-sibling options
A2, then there are 5 non-sibling options
B1, then there are 5 non-sibling options
B2, then there are 5 non-sibling options
C1, then there are 4 non-sibling options
C2, then there are 4 non-sibling options
C3, then there are 4 non-sibling options

The total number of 2 non-sibling options is 5+5+5+5+4+4+4 = 32

The probability is 32/42 = 16/21

Final Answer: E

GMAT assassins aren't born, they're made,
Rich