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

[varun289]

233. 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?

[puneetkhurana2000]

Take the case as 1-2(one sibling each) and 3-4(one sibling each) and 5-6-7(two siblings each) ..so 1-2-3-4 have one sibling each and 5-6-7 have two siblings each.

Total ways 7C2 = 21

Favorable ways 2C1*2C1 + 3C1*2C1 + 3C1*2C1 = 16

Answer is [spoiler]16/21[/spoiler]

[Brent@GMATPrepNow]

233. 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?

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
= [spoiler]16/21[/spoiler]

Cheers,
Brent

[ritind]

@Varun : whats the source of your quant questions. Help will be highly appreciated.

[ritind]

Take the case as 1-2(one sibling each) and 3-4(one sibling each) and 5-6-7(two siblings each) ..so 1-2-3-4 have one sibling each and 5-6-7 have two siblings each.

Total ways 7C2 = 21

Favorable ways 2C1*2C1 + 3C1*2C1 + 3C1*2C1 = 16
Answer is [spoiler]16/21[/spoiler]

Can u explain the highlighted part

[GMATGuruNY]

I posted an explanation here:

https://www.beatthegmat.com/siblings-t66114.html

[puneetkhurana2000]

puneetkhurana2000 wrote:
Take the case as 1-2(one sibling each) and 3-4(one sibling each) and 5-6-7(two siblings each) ..so 1-2-3-4 have one sibling each and 5-6-7 have two siblings each.

Total ways 7C2 = 21

Favorable ways 2C1*2C1 + 3C1*2C1 + 3C1*2C1 = 16
Answer is 16/21

ritind wrote:
Can u explain the highlighted part

2C1*2C1(selecting 1 from (1,2) and selecting 1 from (3,4).
3C1*2C1(selecting 1 from (5,6,7) and selecting 1 from (3,4).
3C1*2C1(selecting 1 from (5,6,7) and selecting 1 from (1,2).

Hope this helps!!!

[ritind]

I posted an explanation here:

https://www.beatthegmat.com/siblings-t66114.html

Thanks a lot Mitch...this is exactly what I was looking for