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# In a room filled with 7 people, 4 people have exactly 1

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01 Jul 2012, 02:15
imhimanshu wrote:
Hi Bunuel,

I have a doubt in approach 1. We have created groups and then we made the selections. i.e selecting one individual from group 1 and another individual from group 2 or select one individual from group 1 and another individual from Group 3 and so on..

However, I would like to ask, can't we dissolve the group and then find the probability.

Lets say, we have total (A,B) (C,D) and (EFG) as groups

Case 1

i.e. Select One individual from Group 1->(A,B) This can be done in 2C1 way and now select the other individual from remaining 5 individuals i.e from C,D,E,F,G - this can be done in 5C1 way.
Therefore, total number of ways 2C1*5C1

Case 2

Similarly, select one sibling from (C,D) and select the other sibling from (A,B,E,F,G) ie total number of ways 2C1*5C1

Case 3

Last Case, select one sibling from (E,F,G) and remaining from (A,B,C,D) i.e. 3C1* 4C1

Whats wrong with this approach. Please clarify.

Thanks
H

Bunuel wrote:
In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4).
As there are 3 people with exactly 2 siblings each: we have one triple of siblings (4-5-6).

Solution #1:
# of selections of 2 out of 7 - $$C^2_7=21$$;
# of selections of 2 people which are not siblings - $$C^1_2*C^1_2$$ (one from first pair of siblings*one from second pair of siblings)+$$C^1_2*C^1_3$$ (one from first pair of siblings*one from triple)+ $$C1^_2*C^1_3$$(one from second pair of siblings*one from triple) $$=4+6+6=16$$.

$$P=\frac{16}{21}$$

I don't understand the red part at all.

As for your solution it's wrong because when for case 1 you select one from {A,B} and then one from {C,D, E, F, G} you can have pair {A,C}. Next, when for case 2 you select one from {C,D} and then one from {A,B,E,F,G} you can have pair {A,C} again. So, you double-count some cases.

Hope it's clear.
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01 Jul 2012, 02:43
Thanks Bunuel, I can see my mistake.

Thanks again.
H
Bunuel wrote:

I don't understand the red part at all.

As for your solution it's wrong because when for case 1 you select one from {A,B} and then one from {C,D, E, F, G} you can have pair {A,C}. Next, when for case 2 you select one from {C,D} and then one from {A,B,E,F,G} you can have pair {A,C} again. So, you double-count some cases.

Hope it's clear.

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Re: In a room filled with 7 people, 4 people [#permalink]

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03 Nov 2012, 16:36
Bunuel wrote:
Archit143 wrote:
In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

5/21
3/7
4/7
5/7
16/21

Merging similar topics. Please refer to the solutions on page 1 (foe example: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html#p645861)

Hope it helps.

BUNEL

Here is how i approached the problem and got the wrong answer can you pls correct me.

Total outcome = 7C2 = 21

Prob of getting exactly 2 sibbling = [4C1 * 3C1] + 3C2 i.e. [ 1 from 4 * 1 from 3] + [ 2 out of 3, since the group of 3 has 2 sibbling]
= [4*3] + 3 =15

prob of getting sibbling = 15 /21

prob of not getting sibbling = 1 - 15/21 = 6/21
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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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12 Nov 2012, 04:38
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Ways to select the those 4 with their sibling?

4/7 x 1/6 = 4/42

Ways to select those 3 with one of their 2 siblings?

3/7 x 2/6 = 6/42

P = 1 - (4/42 + 6/42) = 1 - 10/42 = (42 -10)/42 = 32/42 = 16/21

P = 16/21

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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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13 Dec 2012, 14:08
well, this is how i approached it...

Total 7 people - 1,2,3,4,5,6,7
4 people have 1 sibling - [1,2];[3,4] - 2 single-sibling groups
3 people have 2 siblings - [5,6,7] - 1 two-siblings group

Total ways to select 2 people, 7C2 = 21.
Ways to select only siblings : 2C1( ways to select 1 group from 2 single-sibling groups) + 3C2( ways to select 2 people from 2 sibling-group) = 2+3= 5

Probability that NO siblings selected = 1- 5/21 = 16/21.

Hence, D.

Please let me know if my approach is flawed.
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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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15 May 2013, 00:26
WarriorGmat wrote:
In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

A) 5/21
B) 3/7
C) 4/7
D) 5/7
E) 16/21

If A is a sibling of B, then B is also a sibling of A.
If A is a sibling of B, and B is a sibling of C, that means A is also a sibling of C.

For first four people to have exactly one sibling each means two pairs of siblings.
For last 3 people to have exactly two siblings each means one triplet of siblings.

Now, to select two individuals from a group of 7, total no. of ways it can be done = 7C2 = 21

Cases if siblings are selected : 1. Pair 1
2. Pair 2
3,4,5 = 2 individuals from triplet of siblings = 3C2 = 3 ways.

So, there are 5 cases in which selected individuals are siblings.

Therefore, probability of two individuals selected are NOT sibligs is (21-5)/21 = 16/21
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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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15 May 2013, 00:33
mkdureja wrote:
WarriorGmat wrote:
In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

A) 5/21
B) 3/7
C) 4/7
D) 5/7
E) 16/21

If A is a sibling of B, then B is also a sibling of A.
If A is a sibling of B, and B is a sibling of C, that means A is also a sibling of C.

For first four people to have exactly one sibling each means two pairs of siblings.
For last 3 people to have exactly two siblings each means one triplet of siblings.

Now, to select two individuals from a group of 7, total no. of ways it can be done = 7C2 = 21

Cases if siblings are selected : 1. Pair 1
2. Pair 2
3,4,5 = 2 individuals from triplet of siblings = 3C2 = 3 ways.

So, there are 5 cases in which selected individuals are siblings.

Therefore, probability of two individuals selected are NOT sibligs is (21-5)/21 = 16/21

Hi mkdureja,
Thanks.
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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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15 May 2013, 01:01
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Hi,

Que:
In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

Case 1:
4 people have exactly 1 sibling :
Let A, B, C, D be those 4 people
If A is sibling of B, then B is also a sibling of A.
Basically, they have to exist in pairs
Therefore, these four people compose of two pairs of siblings.

Case 2:
3 people have exactly 2 siblings :
This must obviously be a triplet of siblings.
Let them be E, F, G.
If E is sibling of F, and also is sibling of G., it means F and G are also siblings of each other.
E is sibling of exactly 2 : F and G
F is sibling of exactly 2 : E and G
G is sibling of exactly 2 : E and F
This group has three pairs of siblings.

Now, selecting two individuals out of the group of 7 people has:
1) Two Pairs, as in Case 1.
2) Three Pairs, as in Case 2.
So, 5 cases out of a total of 7C2 = 21 cases.

Hope it clarifies. If you have any further doubt, please point to exactly where you are having a problem understanding it.
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In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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05 Jul 2013, 01:26
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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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05 Jul 2013, 21:07
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In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

It is first important to understand the problem. So let us first assume a specific case.

1. Assume the 7 people are A,B, C, D, E, F and G
2. Assume the 4 people who have 1 sibling are A, B, C and D
3. Let's assume A's sibling is B. Therefore B's sibling is A. Similarly for C and D.
4. So we are left with E, F and G. Each should have exactly 2 siblings.
5. E's siblings will be F and G. So F's siblings will be E and G and G's siblings will be E and F
6. Now look at the general case.
7.Total number of ways of selecting 2 people out of 7 people is 7C2=21
8. Instead of A, B, C and D assume any 4 people. We can see for every such 4 people assumed, there are 2 cases where the selected 2 will be siblings. In the case we assumed they are A and B or C and D. This gives one of the favorable outcomes
9. Or the 2 people selected being siblings may come out of the 3 siblings. The number of favorable outcomes is 3 as we can see in the specific case they are E and F, or F and G or E and G.
10. The total number of favorable outcomes for the selected two being siblings is 2+3=5.
11. The probability that the two selected are siblings is 5/21.
12, Therefore the probability that the two selected are not siblings is 1-5/21= 16/21
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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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14 Jan 2014, 20:53
Quote:
Sure.

We have the following siblings: {1, 2}, {3, 4} and {5, 6, 7}.

Now, in order to select two individuals who are NOT siblings we must select EITHER one from {5, 6, 7} and ANY from {1, 2} or {3, 4} OR one from {1, 2} and another from {3, 4}.

P=2*\frac{3}{7}*\frac{4}{6}+2*\frac{2}{7}*\frac{2}{6}=\frac{4}{7}+\frac{4}{21}=\frac{16}{21}.

3/7 - selecting a sibling from {5, 6, 7}, 4/6 - selecting any from {1, 2} or {3, 4}. Multiplying by 2 since this selection can be don in two ways: the first from {5, 6, 7} and the second from {1, 2} or {3, 4} OR the first from {1, 2} or {3, 4} and the second from {5, 6, 7};

2/7 - selecting a sibling from {1, 2}, 2/6 - selecting a sibling from {3, 4}. Multiplying by 2 since this selection can be don in two ways: the first from {1, 2} and the second from {3, 4} OR the first from {3, 4} and the second from {1, 2}.

Other approaches here: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html#p645861

Hope it's clear.

Hi Bunuel,
Can you please explain to me that why when we use simple probability we need to consider both the cases of selection, if we pick a sibling from the first group of 2 siblings first and then we pick a sibling from the second group of 2 siblings and if we pick a sibling from the second group of 2 siblings first and then we pick a sibling from the first group of 2 siblings.
While if we use combinomatric approach we just count one case i.e. 2C1*2C1 which is for selecting one each from the 2 different 2 sibling groups.
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26 Jan 2014, 10:37
Hi Bunuel
Please correct me where I went wrong.
I assumed the 7 siblings in this way.PFA the image.
So this scenario also satisfies that B.C.F have 2 siblings and A,D,E ,G have 1 sibling each.
Now to select two people that are not siblings :
1)I first select A(1/7) and from rest of the people I can select C,D,E,F,G so the probability of selecting other person that is not sibling is (1/5). And this scenario will be vaild for D,E and G as well So the proability=4*(1/7 * 1/5)
2)Now if I select B (1/7) and from rest of the people I can select ,D,E,F,G so the probability of selecting other person that is not sibling is (1/4). And this scenario will be vaild for B,C,F as well So the proability=3*(1/7 * 1/4)
So Total= 4*(1/7 * 1/5) + 3*(1/7 * 1/4)

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27 Jan 2014, 00:28
282552 wrote:
Hi Bunuel
Please correct me where I went wrong.
I assumed the 7 siblings in this way.PFA the image.
So this scenario also satisfies that B.C.F have 2 siblings and A,D,E ,G have 1 sibling each.
Now to select two people that are not siblings :
1)I first select A(1/7) and from rest of the people I can select C,D,E,F,G so the probability of selecting other person that is not sibling is (1/5). And this scenario will be vaild for D,E and G as well So the proability=4*(1/7 * 1/5)
2)Now if I select B (1/7) and from rest of the people I can select ,D,E,F,G so the probability of selecting other person that is not sibling is (1/4). And this scenario will be vaild for B,C,F as well So the proability=3*(1/7 * 1/4)
So Total= 4*(1/7 * 1/5) + 3*(1/7 * 1/4)

So, as per your diagram (A, B), (B, C), (C, D) are siblings but (A, C), (A, D), and (B, D) are not? How? This is not what question implies.
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27 Jan 2014, 05:33
Bunuel wrote:
282552 wrote:
Hi Bunuel
Please correct me where I went wrong.
I assumed the 7 siblings in this way.PFA the image.
So this scenario also satisfies that B.C.F have 2 siblings and A,D,E ,G have 1 sibling each.
Now to select two people that are not siblings :
1)I first select A(1/7) and from rest of the people I can select C,D,E,F,G so the probability of selecting other person that is not sibling is (1/5). And this scenario will be vaild for D,E and G as well So the proability=4*(1/7 * 1/5)
2)Now if I select B (1/7) and from rest of the people I can select ,D,E,F,G so the probability of selecting other person that is not sibling is (1/4). And this scenario will be vaild for B,C,F as well So the proability=3*(1/7 * 1/4)
So Total= 4*(1/7 * 1/5) + 3*(1/7 * 1/4)

So, as per your diagram (A, B), (B, C), (C, D) are siblings but (A, C), (A, D), and (B, D) are not? How? This is not what question implies.

I still do not understand .The question simply says "In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room". and my diagram meets the given criterion.Also is it necessary that if A&B and B& C are siblings then A&C also need to be siblings.
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27 Jan 2014, 05:36
282552 wrote:
Bunuel wrote:
282552 wrote:
Hi Bunuel
Please correct me where I went wrong.
I assumed the 7 siblings in this way.PFA the image.
So this scenario also satisfies that B.C.F have 2 siblings and A,D,E ,G have 1 sibling each.
Now to select two people that are not siblings :
1)I first select A(1/7) and from rest of the people I can select C,D,E,F,G so the probability of selecting other person that is not sibling is (1/5). And this scenario will be vaild for D,E and G as well So the proability=4*(1/7 * 1/5)
2)Now if I select B (1/7) and from rest of the people I can select ,D,E,F,G so the probability of selecting other person that is not sibling is (1/4). And this scenario will be vaild for B,C,F as well So the proability=3*(1/7 * 1/4)
So Total= 4*(1/7 * 1/5) + 3*(1/7 * 1/4)

So, as per your diagram (A, B), (B, C), (C, D) are siblings but (A, C), (A, D), and (B, D) are not? How? This is not what question implies.

I still do not understand .The question simply says "In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room". and my diagram meets the given criterion.Also is it necessary that if A&B and B& C are siblings then A&C also need to be siblings.

Do you know what a sibling mean? How can (A, B), (B, C), (C, D) be BROTHERS, and (A, C), (A, D), and (B, D) not to be?
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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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29 Jan 2014, 10:33
Hi,

Can someone please explain to me that why when we use simple probability we need to consider both the cases of selection i.e. if we pick a sibling from the first group of 2 siblings(A,B) first and then we pick a sibling from the second group of 2 siblings(C,D) and if we pick a sibling from the second group of 2 siblings(C,D) first and then we pick a sibling from the first group of 2 siblings(A,B)
While if we use combinomatric approach we just count one case i.e. 2C1*2C1 which is for selecting one each from the 2 different 2 sibling groups

How is A,C and C,A different if we just have to check that whether the two selected are siblings or not? And if it makes a difference why we did not you permutation instead of combinations?
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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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18 May 2014, 23:49
$$2*\frac{1}{7}*\frac{5}{6}+2*\frac{1}{7}*\frac{5}{6}+3*\frac{1}{7}*\frac{4}{6}$$
$$\frac{32}{42} = \frac{16}{21}$$
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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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08 Oct 2014, 03:16
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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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08 Oct 2014, 03:43
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Expert's post
keyrun wrote:

Shen you select groups 1 and 2, you have 3*2 choices, when groups 1 and 3 you have 3*2 choices and when groups 2 and 3 you have 2*2 choices: total = 6 + 6 + 4.
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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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29 Jun 2015, 23:43
Hi everyone

I was stuck with this question b/w the no of pairs of siblings and the answer

I thought i should post this.

May be it helps someone out dr.

Happy learning
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