Probability

[sparkle6]

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


[spoiler]Answer: E. Can someone pleas explain using combination theory?[/spoiler]

[knight247]

We have a group of 7 people. After careful analysis of the problem we can figure out the following...
We have one sibling trio x1 x2 x3
we have one sibling pair a1 and a2
we have another sibling pair b1 and b2

Out of seven when we pull out a person we are obviously not going to replace him. Also when two people are pulled out we have two possible outcomes i.e. either both are siblings of each other or both are not siblings of each other. So the two outcomes are complementary meaning either one can occur but both can't occur. So we have a probability problem without replacement with complementary outcomes.

P(Both are siblings of each other)+P(Both are not siblings of each other)=1

First we'll find the answer using the direct method. For this we have to consider 3 different scenarios.

1.Our first pick is from x1 x2 and x3 and the second pick can be from any 4 of the remaining 6 so we have
3/7*4/6=6/21

2. Out first pick is from a1 a2 and our second pick can be from any 5 of the remaining 6 so we have
2/7*5/6=5/21

3. This scenario is same as 2 so 5/21

Adding all three we have 6/21+5/21+5/21=16/21


Next, we can solve by finding the probability where the two ppl picked will always be siblings
Again we consider 3 different scenarios

1.Our first pick is from the sibling group x1 x2 x3 and the second pick has to be from the same group
So we have. 3/7*2/6=1/7=3/21

2. Our first pick is from the sibling group a1 a2 and the second pick also has to be from the same group. So we have 2/7*1/6=1/21

3. Same as scenario 2 so 1/21

Adding all the three ways we have 1/21+1/21+3/21=5/21 .This is the probability that in the two picks both will be siblings. So 1-5/21=16/21 which is the probability that in the two picks neither will be siblings. Hope this clarifies everything.

[Brent@GMATPrepNow]

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


[spoiler]Answer: E. Can someone pleas explain using combination theory?[/spoiler]

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 = E

Cheers,
Brent