Wanted to bump this question up mainly because this question is great for back-up method (thanks ron).

Method 3 to doing this problem - just list them!

First, total number of ways of picking 2 people out of 7 = (7*6)/2! = 21.

Next, the number of ways to pick a pair of sibling:

Use a letter to denote each person in the group:

A-B = first 2-sibling group
C-D= second 2-sibling group
E-F-G = 3-sibling group

So your pairs are:

AB
CD
EF
FG
EG

(remember here the positions are interchangeable... so AB and BA are the same pair)

5 total ways to pick two people who are siblings, so you probability of picking a sibling is 5/21.

Applying the 1-x rule: 1 - (5/21) = 16/21 probability of NOT sibling.

Bravo MGMAT - very nice question.

Glad you liked it James. Maybe we should add Ron's note to the CAT explanation.

4/7 you pick someone with 1 sibling. 3/7 you pick someone with 2 siblings.

If you pick a 1 sibling person, he has one sibling in the deck of 6 remaining people: 4/7 x 1/6 = 4/42 = 2/21

If you pick a 2 sibling person: 3/7 x 2/6 = 6/42 = 3/21

Add them: 5/21 says you get a sibling pair.

1 - 5/21 = 16/21 says you don't.

Nice work everyone! Thank you to our contributors.

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Jamie Nelson
ManhattanGMAT Instructor

Can't we solve this problem as -

P(no siblings selected) = 1 - P( siblings selected)

P( siblings selected) = P(siblings from group 1) + P (Siblings from group 2) + P (siblings from group 3)

groups - are the three groups based on having two and three sibling pairs as said above in the posts

= 2C2/7C2 + 2C2/7C2 + 3C2/7C2
= 5/21

Then P(no siblings) = 1- 5/21 = 16/21

Please point out if I missed anything?

Can't we solve this problem as -

P(no siblings selected) = 1 - P( siblings selected)

P( siblings selected) = P(siblings from group 1) + P (Siblings from group 2) + P (siblings from group 3)

groups - are the three groups based on having two and three sibling pairs as said above in the posts

= 2C2/7C2 + 2C2/7C2 + 3C2/7C2
= 5/21

Then P(no siblings) = 1- 5/21 = 16/21

Please point out if I missed anything?

Totally appropriate to do it this way, and you did it perfectly!

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Tim Sanders
Manhattan GMAT Instructor

Conditional Probability for Grade XII math question?
I had this question on a quiz but it was never taken up and I'm curious as to what the answer is. This is the question:

In a group of 60 students, 25 have blonde hair, 15 have blue eyes, and 33 have neither. Determine the probability of a student having both blond hair and blue eyes.

So as my answer I used the non-mutually exclusive formula and wrote:
P(A|B) = P(A) + P(B) -33
=25+15+33
=7

The answer was marked incorrect. I was wondering if I even used the right formula. I didn't think it was conditional probability but I tried it anyway:

P(A|B) = 25/15
=1.67

Is this right? Because you can't really have 1.67 people...
The only other thing I can think of is using the mutually exclusive formula which would be:

P(A∪B) = 25 + 15
=40
But I really don't think that's right... If you know what I'm doing wrong or which solution is indeed correct, could you please explain to me how you came to that conclusion? Thank you very much.

This question is just far enough beyond the reach of what we've ever seen on the GMAT that I won't answer it here on the forum. In general though, please note that our double set matrix approach to overlapping sets questions is always preferable to and more versatile than using a formula you've memorized (conditional probability, PIE, etc.)..

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Tim Sanders
Manhattan GMAT Instructor