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Manager  G
Joined: 07 Jun 2017
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In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

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12 00:00

Difficulty:   75% (hard)

Question Stats: 55% (02:13) correct 45% (02:31) wrong based on 95 sessions

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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. $$\frac{17}{110}$$
Math Expert V
Joined: 02 Sep 2009
Posts: 59739
In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

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4
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nkmungila wrote:
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. $$\frac{17}{110}$$

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.

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Math Expert V
Joined: 02 Sep 2009
Posts: 59739
Re: In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

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Bunuel wrote:
nkmungila wrote:
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. $$\frac{17}{110}$$

5 have 4 siblings each implies that there is one group of 5 siblings {1 - 2 - 3 - 4 - 5};
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.

Similar question: https://gmatclub.com/forum/in-a-room-fi ... 87550.html
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In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

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nkmungila wrote:
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. $$\frac{17}{110}$$

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
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Re: In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

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nkmungila wrote:
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. $$\frac{17}{110}$$

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.

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Re: In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

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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.
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Re: In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

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chesstitans wrote:
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.

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.
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Re: In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

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DavidTutorexamPAL wrote:
chesstitans wrote:
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.

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.

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.
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Posts: 9880
Location: Pune, India
Re: In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

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chesstitans wrote:
DavidTutorexamPAL wrote:
chesstitans wrote:
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.

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.

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.

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

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Re: In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

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