GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Oct 2019, 11:03

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# There are 9 people in the room. There are two pairs of

Author Message
TAGS:

### Hide Tags

Manager
Joined: 15 Aug 2013
Posts: 228
Re: There are 9 people in the room. There are two pairs of  [#permalink]

### Show Tags

21 Apr 2014, 18:53
Bunuel wrote:
russ9 wrote:
There are 9 people in the room. There are two pairs of siblings within that group. If we choose two people, what is the probability that they will not be siblings?

Can someone please explain where i'm going wrong?

I'm using the approach: No Sibling Pair = 1 - Probability of sibling pair.

P = $$\frac{(9C4)(1C1)}{(9C2)}$$ -- why is this wrong? Is this equation not implying that at first, I have the probability of choose 4 people out of 9, then I can only pick 1 so my top is the favorable solution. My denominator is the total number of outcomes.

The numerator is wrong. We are not choosing 4 people, we are choosing 2.

How many ways to choose a sibling pair from 9 people where there are two pairs of siblings (X, X, Y, Y, A, B, C, D, E)? Only two ways XX and YY.

Hence, P = 1 - 2/(9C2) = 1 - 1/18 = 1/17.

Hi Bunuel,

I am reading what you're writing but I just cant grasp it. How is this different than the other problems where we try to keep the numerator and denominator in the same units, i.e: people and not pairs. Is P = $$\frac{(9C4)(1C1)}{(9C2)}$$ not saying that i want to pick ANY member out of the 4 siblings and then I can only choose 1 member after choosing that ANY member?
Intern
Joined: 30 Oct 2013
Posts: 3
Re: There are 9 people in the room. There are two pairs of  [#permalink]

### Show Tags

09 Jun 2014, 04:03
There are 9 people in the room. There are two pairs of siblings within that group. If we choose two people, what is the probability that they will not be siblings?

I need to choose two people from a group of 9 . I will choose 1 the probablitiy is 100% (9/9), then I have to choose the second person in such a manner that he is not the sibling of the 1 I chose, therefor amongst the 8 I am still left with, the probability becomes 7/8 (excluding the one sibling of the first). Please correct my approach and preferably donot provide me with an alternative solution. It would be of great help.
Math Expert
Joined: 02 Sep 2009
Posts: 58381
Re: There are 9 people in the room. There are two pairs of  [#permalink]

### Show Tags

09 Jun 2014, 09:10
ankitarihaan wrote:
There are 9 people in the room. There are two pairs of siblings within that group. If we choose two people, what is the probability that they will not be siblings?

I need to choose two people from a group of 9 . I will choose 1 the probablitiy is 100% (9/9), then I have to choose the second person in such a manner that he is not the sibling of the 1 I chose, therefor amongst the 8 I am still left with, the probability becomes 7/8 (excluding the one sibling of the first). Please correct my approach and preferably donot provide me with an alternative solution. It would be of great help.

The point is that if you choose a person which has no sibling then the probability that the second person won't be his/her sibling would be 1. That's why this approach does not work.

Correct approach is here: there-are-9-people-in-the-room-there-are-two-pairs-of-58609.html#p1357845

Hope it helps.
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 13208
Re: There are 9 people in the room. There are two pairs of  [#permalink]

### Show Tags

16 Sep 2019, 11:17
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: There are 9 people in the room. There are two pairs of   [#permalink] 16 Sep 2019, 11:17

Go to page   Previous    1   2   [ 24 posts ]

Display posts from previous: Sort by