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In a room filled with 7 people, 4 people have exactly 1 sibling in the

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

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New post 15 May 2013, 02:01
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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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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 05 Jul 2013, 22:07
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reply2spg 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


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 sibling in the  [#permalink]

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New post 26 Jan 2014, 11:37
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)

Which does not lead to a correct answer.
Please HELP!!!!!!!!!!!
And please pardon my bad drawing skills :(
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 27 Jan 2014, 01:28
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)

Which does not lead to a correct answer.
Please HELP!!!!!!!!!!!
And please pardon my bad drawing skills :(


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

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New post 27 Jan 2014, 06:33
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)

Which does not lead to a correct answer.
Please HELP!!!!!!!!!!!
And please pardon my bad drawing skills :(


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

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New post 27 Jan 2014, 06:36
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)

Which does not lead to a correct answer.
Please HELP!!!!!!!!!!!
And please pardon my bad drawing skills :(


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 sibling in the  [#permalink]

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New post 19 May 2014, 00:49
\(2*\frac{1}{7}*\frac{5}{6}+2*\frac{1}{7}*\frac{5}{6}+3*\frac{1}{7}*\frac{4}{6}\)
\(\frac{32}{42} = \frac{16}{21}\)
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 08 Oct 2014, 04:16
Can anyone please help me on where I went wrong?
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 30 Jun 2015, 00:43
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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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 08 Jul 2015, 10:00
I am posting a visual solution.

To begin with, there are 7!/2!*5! = 21 total possibile pairings.

After finding the no-sibling pairings (see image below) we end up with 16/21.
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 10 Jul 2015, 20:46
x = P(1-sibling person) = 4*p(1-sibling person)
y = P(2-sibling person) = 3*p(2-sibling person)

p(1-sibling person)=1/7*1/6 (1/6 because we have 1 sibling in the 6 remaining people)
p(2-sibling person)=1/7*2/6 (2/6 because we have 2 siblings in the 6 remaining people)

Answer = 1-x-y = 16/21

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

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New post 11 Dec 2015, 03:50
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reply2spg 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


This is how I solved.

Case 1: The group of 4 have exactly 1 sibling. So maximum 2 people can be picked from that group who are not siblings.
Case 2: The group of 3 has 2 siblings each. So maximum only 1 person from the group can be chosen.
Total number of ways in which we can pick people is 7C2 = 21.


Case 1 -
\(2C1 * 2C1\) ways

Case 2 -
\(3C1 * 4C1\) ways

Probability -
\(\frac{2C1 * 2C1 + 4C1 * 3C1}{21} = \frac{16}{21}\)

Experts, please confirm if my approach is correct :)

+Kudos if this helped!
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 12 Aug 2017, 11:49
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lagomez wrote:
You have people 1-2-3-4-5-6-7

4 have exactly 1 sibling which can mean:
1-2 are siblings
3-4 are siblings

3 have exactly 2 siblings which can mean:
5-6-7 are siblings

Let's start with the group of 4:
The probability of picking 1 is (1/7)
The probability of not getting a sibling pair is (5/6) because the only other sibling is 2
Therefore, if 1 is selected first the probability of not getting a sibling pair is 5/42
Multiply that by 4 because the probability is the same whether you start with 1-2-3-4 so you get 20/42 for the first 4 people

Now let's go to the group of 3:
The probability of picking 5 is (1/7)
The probability of not getting a sibling pair is (4/6) which the non-sibling pair is 1-2-3-4
Therefore the probability is 4/42. Multiply that probability by 3, which represent 5-6-7 so the probability is 12/42

Now you have two probabilities: 12/42 and 20/42
add both and you get 32/42 or 16/21



It's easier to use the complement. Not sure why people are trying so hard here to use the choose formula when it makes the question 1000X harder

4/7 chance to pick a person with 1 sibling * 1/6 chance you get a sibling or 4/42

3/7 change to pick a 2 person sibling * 2/6 or 6/42

Total of 10/42 of getting a sibling.

We get to use the simple or rule not the generalized since the things are mutually exclusive. And we aren't drawing twice just once. So either a 4 person is picked first or 3 person.

So that simplifies to 5/21 then 1 minus this is 16/21 for the odds that they are not siblings. 5/21 they are.
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 11 Aug 2018, 05:28
I have used alphabet analogy to solve this Q.

Consider three groups, each containing same alphabets.

Group 1: AA
Group 2: BB
Group 3: CCC

We need to select 2 distinct alphabets.

TOTAL OUTCOMES:
ways of selecting 2 alphabets out of 7 without any restriction is 7C2= 21

FAVORABLE OUTCOMES:
based on restriction of selecting distinct alphabets, we have following possibilities:

1 each from G1&G2: 2C1 x 2C1= 4
1 each from G1&G3: 2C1 x 3C1= 6
1 each from G2&G3: 2C1 x 3C2= 6
Total favorable outcomes: 16

Probability: 16/21

Ans E

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

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New post 03 Oct 2018, 17:53
PiyushK wrote:
\(2*\frac{1}{7}*\frac{5}{6}+2*\frac{1}{7}*\frac{5}{6}+3*\frac{1}{7}*\frac{4}{6}\)
\(\frac{32}{42} = \frac{16}{21}\)


this one made the most sense to me.

We have two groups of ppl:

A B C D
E F G

starting from A, chance of picking the first person:
1/7
after A, the chance of not picking a sibling:
5/6

Repeat this for B, C, and D
2*(1/7)(5/6)

If we pick E F G first
then, we know the first chance is 1/7, but second chance is 4/6
hence (1/7)(4/6)
repeat this for E F G
hence 3x

so total should
2*(1/7)(5/6)+2*(1/7)(5/6)+3(1/7)(4/6)
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 10 Feb 2019, 19:31
I think this question can be a lot easier if we realize that within the 4 people (A,B,C,D) who has exactly 1 sibling, there are two 'sets' of siblings -- any selection two people within the group of 2 could be siblings with each other. The three people(E,F,G) with exactly 2 siblings, are siblings with each other.

So instead of breaking the 7 people into three groups, we can solve it by dividing the people into two groups, making up 3 cases.

Case 1: ABCD, Pick 2
4C2 = 6. But of the 6 potential sets of siblings, we already know that there are siblings - so subtract 2, we have 4.

Case 2: EFG, pick 2
3C2=3. Again, we already know that the three people are siblings with each other, so all three potential sets will be siblings - so subtract 2, we get 0 from this group.

Case 3: Pick one from ABCD, pick another from EFG.
No restrictions here, people from 2 sibling group are not the same people from 1 sibling group.
4C1+3C1=12

Non-sibling selections: 4+0+12=16
Total selections: 7C2=21
Part/All = 16/21
Answer E
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 13 May 2019, 11:36
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reply2spg 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


First we need to recognize that the given information tells us that the 7 people consist of:
- a sibling trio
- a sibling pair
- and another sibling pair

Let's use counting techniques to answer this question

For this question, it's easier to find the complement.
So P(not siblings) = 1 - P(they are siblings)

P(they are siblings) = [# of ways to select 2 siblings] / [total # of ways to select 2 people]

# of ways to select 2 siblings
Case a) 2 siblings from the sibling trio: from these 3 siblings, we can select 2 siblings in 3C2 ways (3 ways)
Case b) 2 siblings from first sibling pair: we can select 2 siblings in 2C2 ways (1 way)
Case c) 2 siblings from second sibling pair: we can select 2 siblings in 2C2 ways (1 way)

So, total number of ways to select 2 siblings = 3+1+1 = 5

total # of ways to select 2 people
We have 7 people and we want to select 2 of them
We can accomplish this in 7C2 ways (21 ways)

So, P(they are siblings) = 5/21

This means P(not siblings) = 1 - 5/21
= 16/21

Answer: E

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

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New post 03 Jun 2019, 13:56
I calculated the probability they would be siblings:

(4/7 *1/6) + (3/7 * 2/6) = 10/42 = 5/21 and then i subtracted this from 1 to get 16/21.
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 30 Jun 2019, 07:27
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 (5-6-7).

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) \(=4+6+6=16\).

\(P=\frac{16}{21}\)

Solution #2:
# of selections of 2 out of 7 - \(C^2_7=21\);
# of selections of 2 siblings - \(C^2_3+C^2_2+C^2_2=3+1+1=5\);

\(P=1-\frac{5}{21}=\frac{16}{21}\).

Solution #3:
\(P=2*\frac{3}{7}*\frac{4}{6}+2*\frac{2}{7}*\frac{2}{6}=\frac{4}{7}+\frac{4}{21}=\frac{16}{21}\).

Answer: E.




Bunuel

in solution 1, why are we taking it to 3 sections, after combination from both the groups i.e \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings),
why cant we select one from triple and then any one from the rest for as they wont be siblings to each other.

So i took it like
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) + \(C^1_3*C^1_4\) (one from triple*one from rest)

What is wrong in here?? I'm really confused.
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