In a group of 11 members, 5 have 4 siblings each and 6 have one

Question

In a group of 11 members, 5 have 4 siblings each and 6 have one sibling each. If two members are picked randomly, what is the probability that they are siblings?

A.  \frac{16}{55}

B.  \frac{13}{55}

C.  \frac{10}{55}

D.  \frac{6}{11}

E.  \frac{17}{110}

Bunuel · Accepted Answer

5 have 4 siblings each implies that there is one group of 5 siblings {1 - 2 - 3 - 4 - 5} (notice that each of these 5 have 4 siblings);
6 have one sibling each implies that there are three groups of 2 siblings {6 - 7}, {8 - 9}, {10 - 11}.

The total number of ways to choose two people out of 11 is 11C2 = 11!/(2!9!) = 55.

In order to get two siblings we can choose any two from {1 - 2 - 3 - 4 - 5}, so 5C2 = 10 cases or any of the three sibling pairs: {6 - 7}, {8 - 9} or {10 - 11}, so 3 cases.

P = (10 + 3)/55 = 13/55.

Answer: B.

Bunuel · Answer

Similar questions to practice: https://gmatclub.com/forum/in-a-room-fi ... 87550.html https://gmatclub.com/forum/in-a-group-o ... 50327.html

chetan2u · Answer

Hi....

5 have 4 siblings each means these 5 are siblings
6have 1 sibling each means there are 3 pair of siblings.

So two ways to solve..

1) finding prob of picking siblings..
As also explained above by bunuel...
Picking 2 out of 5 siblings =5C2=10
Picking 2 of the two siblings=2C2=1, three such pairs so 3 ways..
Total 10+3=13..
Ways to pick up 2 out of 11=11C2=11*10/2=55

Prob =13/55

2) prob of not picking siblings
Pick up 1 of these 5siblings and SECOND can be any of remaining 6, so 5*6=30
Pick up 1 of the two siblings and SECOND can be any of 11-2 or 9, so 2*9..
Three such pairs=2*9*3=54

Total 30+54=84..
Way to pick up 2 =11*10
So prob of not picking=84/11*10=84/110=42/55
So prob of picking=1-42/55=13/55

B

ScottTargetTestPrep · Answer

Of the 5 people who have 4 siblings each, they must be each others’ siblings. Of the 6 people who have 1 sibling each, there must be 3 pairs of people who are siblings of each another.

When 2 people are picked randomly, the probability that they are 2 people from the 5-people siblings is:

5/11 x 4/10 = 20/110

The probability that they are 2 people from any of the 3 pairs of siblings is:

(2/11 x 1/10) x 3 = 6/110

Thus, the overall probability is 20/110 + 6/110 = 26/110 = 13/55.

Answer: B

chesstitans · Answer

I am really bad at probability. I still get confused with A. I even get paranoid, and comes up with something else.
This question is extremely important, but I got a wrong answer.

DavidTutorexamPAL · Answer

Can you be a bit more specific about what you find difficult?

In this question, the 'trick' is to realize that there are 4 groups - 1 containing 5 people and 3 containing two people.
Once you've realized that, the solution method is fairly standard - just choose people from your groups and divide by the total possibilities.

So the hard part of the question is selecting the right 'choice model'.
Which model should you use? It's best to learn this via experience - by solving lots of questions in the field.

chesstitans · Answer

the problem is not which math model I use, but the logical sense that I try to comprehend from applying the models. I feel frustrated.

KarishmaB · Answer

It is important to realise what each statement means:

5 have 4 siblings each - There are a total of 11 people. Say A has 4 siblings so it means we have a group of 5 brother-sister. Each of those 5 brother-sister has exactly 4 siblings. So we have 5 people with 4 siblings each.
6 have one sibling each - Say B has a sibling C. So C also has sibling B. So we have 2 people who have a sibling each. Similarly, we will have 2 more pairs of 2 siblings each. In all we have 3 pairs of 2 siblings each.

So here are sibling groups:
A-P-Q-R-S
B-C
D-E
F-G

Probability of selecting a sibling pair:
We could select one of 5 with a probability of 5/11.
We could then select a sibling for her with a probability of 4/10
Probability = (5/11)*(4/10) = 20/110

OR

We could select one of the other 6 with a probability of 6/11
Her sibling is 1 of the leftover 10 so we can select him with a probability of 1/10
Probability = (6/11)*(1/10) = 6/110

Total Probability = 26/110 = 13/55

Answer (B)

100mitra · Answer

Correct option B

1. You need to select 2 memeber from 11 
I.e 11C2 - (11x10)/2 = 55

2. Their are 2 condition : 
(Condition 1)- 5 member has 4 siblings = 5C2 : (5x4)/2 = 10
Or 
(Condition 2)- 6 member has each siblings =(2C2) for each 3 condition. 1+1+1 = 3

Since there are 2 condition are independent for probability, we need to add.
(10+3) / 55 = (13/55)
The winner B

Additional,
- if we have to select 1 from each condition, which is like depending on each other, you need to multiply
(13x3) / 55 = 39/55

Posted from my mobile device

rvgmat12 · Answer

11 people

5 people = 1 group (each one has 4 siblings)

3 other groups (each one has one sibling)

(5C2+ 3)/11C2 = 13/55

B

sashankmvvv · Answer

Don't we need to make cases 6 siblings, because we don't know what there are possibilities, right? How do we know that a 6C2 won't be the correct answer? Because the number of siblings can be AB, AC,AD, etc.

Bunuel · Answer

Your doubt is not entirely clear. However, if you mean that the six having one sibling each could be interpreted as forming a group like {1, 2, 3, 4, 5, 6}, where each person is a sibling of all the others, then this does not qualify because in that case each person would have five siblings, not one. Please refer to the discussion above for correct interpretation and solutions.

In a group of 11 members, 5 have 4 siblings each and 6 have one

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In a group of 11 members, 5 have 4 siblings each and 6 have one

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