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

Question

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?

A. 1/18
B. 7/18
C. 9/18
D. 13/18
E. 17/18

eschn3am · Accepted Answer

9 people are standing in a room.
Included are 2 pairs of siblings (so 4 people, each with on sibling)
What is the probability of picking two people that are NOT siblings.

Start with 1, because that's 100% probability. From here we'll subtract the chances of getting a pair of siblings. What we have left is the probability of choosing 2 people who are NOT siblings.

Now we start our choosing. We have 9 people to choose from for our 1st choice. Out of these 9 people we must choose one of the 4 with siblings. If we choose one of the 5 people without siblings then our second choice won't matter at all because they have no relatives we can pick up.

https://gmatclub.com/cgi-bin/mimetex.cgi?formdata=%5Cfrac%7B4%7D%7B9%7D  = probability of picking 1 of 4 people with siblings from a group of 9

Now we have to find how many ways we can choose that persons sibling once we have chosen one of the 4. We have 8 people left to choose from and since there are pairs of siblings, each of the 4 people only has ONE sibling. Thus, there is only 1 person we can choose out of the remaining 8.

https://gmatclub.com/cgi-bin/mimetex.cgi?formdata=%5Cfrac%7B1%7D%7B8%7D  probability of picking the sibling of the first person we selected

Put it all together:  https://gmatclub.com/cgi-bin/mimetex.cgi?formdata=1-%28%5Cfrac%7B4%7D%7B9%7D*%5Cfrac%7B1%7D%7B8%7D%29%3D+1-%5Cfrac%7B4%7D%7B72%7D%3D%5Cfrac%7B68%7D%7B72%7D+%3D+%5Cfrac%7B17%7D%7B18%7D

maratikus · Answer

here is another solution. The number of ways to pick 2 people from a group of 9 is equal to 9*(9-1)/2 = 36

there are only 2 pairs of siblings in the room. Therefore, probability of choosing a pair of siblings is 2/36 = 1/18. Probability of choosing a pair who are not siblings is 1-1/18 = 17/18

eschn3am · Answer

https://gmatclub.com/cgi-bin/mimetex.cgi?formdata=1-%28%5Cfrac%7B4%7D%7B9%7D*%5Cfrac%7B1%7D%7B8%7D%29%3D+1-%5Cfrac%7B4%7D%7B72%7D%3D%5Cfrac%7B68%7D%7B72%7D+%3D+%5Cfrac%7B17%7D%7B18%7D

walker · Answer

another ways: 1. p=\frac{C^9_2-C^2_1}{C^9_2}=1-\frac{2}{36}=\frac{17}{18} or 2. p=\frac{P^9_2-P^2_1*P^2_2}{P^9_2}=1-\frac{4}{72}=\frac{17}{18} or 3. p=\frac49*\frac78+\frac59*\frac88=\frac{68}{72}=\frac{17}{18} probability questions always have a few ways :boxer2

bmwhype2, thanks for your question +1

x2suresh · Answer

= 1- (2C2*2/9C2) = 1-1/18 = 17/18

sanjay_gmat · Answer

Method 1  : total number of ways of choosing 2 people out of 9 = 9C2 = 36

probability of choosing two people so that they are siblings = 2C1 = 2 { since there are two pairs of siblings }

required probability = (36 -2) /36 = 17/18.

Method 2  : total number of ways = 36

let say the nine people that we have are A, B, C, D, E, S11, S12, S21, S22 where (S11, S12) and (S21, S22) are sibling pairs.

number of ways one can choose two people is :

choose 2 from A, B, C, D, E . Number of ways = 5C2 = 10.

choose 1 from A, B, C, D, E AND 1 from (S11, S12, S21, S22) in 5*4 = 20 ways

choose 1 from (S11, S12) and 1 from (S21, S22) in 4 ways .

total number of ways = 34

probability = 34/36 = 17/18

srivas · Answer

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?

Soln: probability that they will not be siblings
= 1 - Probability that they will be siblings
= 1 - (4/9 * 1/8)
= 1 - (1/18)
= 17/18

dkj1984 · Answer

Hi Walker,

With reference to your solution for the married couples problem where in we have to select 4 people out of 6 couples such that none is married to each other,can this problem be tackled in the same way?

How will we arrive at the solution?

Thanks!

regards,

walker · Answer

You can use the same approach but it's important to note that not all people have siblings.

C^5_2  - the number of options to choose 2 people out of 5 people without siblings
 C^5_1*C^2_1*C^2_1  - the number of options to choose 1 person out of 5 without siblings, choose 1 pair of siblings out of 2, choose 1 person out of the pair.
 C^2_1*C^2_1  - the number of options to choose 1 person out of 2 for each pair of siblings.

p = \frac{C^5_2+C^5_1*C^2_1*C^2_1+C^2_1*C^2_1}{C^9_2} = \frac{10+20+4}{36} = \frac{34}{36} = \frac{17}{18}

Maxirosario2012 · Answer

My approach :idea:

: 1- p(1 minus the probability that the people we pick will be siblings) We have 9 people: A1 - A2 - B1 - B2 - C - D - E - F - G Probability of picking siblings: If we first pick A1, Probability of picking A1 p=\frac{1}{9}\frac (we pick 1 between 9) Then, probability of picking the sibling of A1, A2, after having picked A1: p=\frac{1}{8}\frac (we pick 1 between the remainder 8 people) * \frac{1}{9}\frac (conditional probability of picking A2 after having picked A1) = \frac{1}{72}\frac But, in addition, we could have picked A2 and then A1. Then: p=\frac{1}{72}\frac *2 = \frac{2}{72}\frac = \frac{1}{36}\frac The same if we picked first B1 and secondly B2 or ("or" means "+") if we first picked B2 and secondly B1. Then, the total probability of picking siblings is: p=\frac{1}{36}\frac * 2 = \frac{2}{36}\frac = \frac{1}{18}\frac Then, 1 - p = \frac{18}{18}\frac - \frac{1}{18}\frac = \frac{17}{18}\frac Is right this logic? I am understanding the problem correctly :?:

Bunuel · Answer

Absolutely, though not the most elegant way of solving.

Maxirosario2012 · Answer

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?
The reversal probability approach is the most efficient way to solve. But I explore the other methods just for practice. Tell me your opinions, I have a little doubt on the reversal combinatorial approach.

Probability approach:
   
XXXXXAABB

P= \frac{6}{9} * \frac{1}{2}*2 +\frac{5}{9}*\frac{4}{8} = \frac{12}{18} + \frac{5}{18} = \frac{17}{18}

\frac{6}{9}*\frac{1}{2}*2 : we choose 6 people out of 9 (XXX XXA, 6/9) and then 1 people out of the other 2 (BB 1/2). Then multiply by 2 because we choose again 6 people out of 9 (XXXXXB) and then 1 people out of the other 2 (AA).

\frac{5}{9}*\frac{4}{8} : then we focus only on the other 5 that are not siblings (XXXXX). We choose this 5 people (XXXXX) out of all 9 (then 5/9) and we combine this one with the remaining 4 of the group (XXXX) of the remaining 8.

Reversal Probability approach:   
P = 1-q (q=probability of choosing one sibling)

q= \frac{2}{9} * \frac{1}{8} + \frac{2}{9} * \frac{1}{8} = \frac{4}{72} = \frac{1}{18} 
 P = 1-q = \frac{18}{18} - \frac{1}{18} = \frac{17}{18}

\frac{2}{9} : probability of picking the person A from AABBXXXXX
 \frac{1}{8} : probability of picking the other A from the remaining ABBXXXXX

Then we add the probability of picking B from AABBXXXXX and probability of picking again B from AABXXXXX

Combinatorial approach:

P= \frac{{C^6_1 * C^2_1 + C^6_1 * C^2_1 + C^5_2}}{C^9_2}= \frac{{6*2 + 6*2 + 10}}{36} = \frac{34}{36} = \frac{17}{18}

C^6_1 * C^2_1 : we choose one people from AXXXXX and combine with one people of BB
 C^6_1 * C^2_1 : we pick one people of BXXXXX and combine it with one people of AA 
 C^5_2 : we choose 2 people from XXXXX
 C^9_2 : total combinations of 2 from 9.

Reversal Combinatorial approach:

q= \frac{{C^2_1 + C^2_1}}{C^9_2}= \frac{{2 + 2}}{72} =\frac{4}{72} = \frac{1}{18} 
 P = 1-q = \frac{18}{18} - \frac{1}{18} = \frac{17}{18}

C^2_1 : we choose 2 persons from the sibling (AA)
 C^2_1 : we choose 2 persons from the sibling (BB)

russ9 · Answer

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.

Bunuel · Answer

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 = 17/18.

ankitarihaan · Answer

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.

Bunuel · Answer

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.

vernondcruz123 · Answer

You could also think of it like this:

Sibling Pair 1: A1, A1
Sibling Pair 2: B1, B2

2 People Selected but both not siblings would mean, the sum of these cases
Case 1: A1 selected but not A2 = 1/9 * 7/8 
Case 2: A2 selected but not A1 = 1/9 * 7/8
Case 3: B1 selected but not B2 = 1/9 * 7/8 
Case 4: B2 selected but not B1 = 1/9 * 7/8
Case 5: Any one other than the 4 siblings selected on the first go followed by anyone from the remaining 8 selected on the second go = 5/9 * 1

FInal = 4 * (1/9 * 7/8) + 5/9 = 17/18

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