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

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

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Updated on: 30 Jun 2019, 10:07
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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

Originally posted by reply2spg on 01 Nov 2009, 10:23.
Last edited by Bunuel on 30 Jun 2019, 10:07, edited 2 times in total.
Renamed the topic.
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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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01 Nov 2009, 10:51
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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

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

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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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01 Nov 2009, 19:28
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12
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

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
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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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02 Dec 2009, 21:27
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For me it is pretty straightforward:

two mutually exclusive events:

1) we have 4/7 probability of getting 1-sibling person and 5/6 probability of not getting his or her sibling.
2) we have 3/7 probability of getting 2-siblings person and 4/6 probability of not getting his or her sibling.

p=4/7*5/6+3/7*4/6 = 16/21

- Probability
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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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02 Dec 2009, 22:50
1
2
I also get the same result 16/21, but my approach is a bit different.
There are 7C2 =21 ways to select 2 people in a group of 7.
There are 1+1+3 = 5 ways to select 2 people who are siblings.
So the probability of selecting 2 people who are not siblings is : 1-5/1 =16/21

@ Walker: I also tried to solve the problem in the way that you suggested, but i can't get myself clear why the two events mentioned are mutually exclusive.
I thought the following events are mutually exclusive for this problem, and i arrived with the following solution
1) 1 person from 2-sibling group and 1 person from 1-sibling group, which is the same with your second event:
we have 3/7 probability of getting 2-siblings person and 4/6 probability of not getting his or her sibling.
2) No one from the 2-sibling group.
we have 4/7 probability of getting 1-sibling person and 2/6 probability of not getting his or her sibling and not within one of 2-sibling group.
p = 3/7*4/6 + 4/7*2/6 = 20/42 (this is not within the answers, so i guess it is wrong, but could you please help me where I got wrong with this). Many thanks.
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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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03 Dec 2009, 05:58
Fiven wrote:
@ Walker: I also tried to solve the problem in the way that you suggested, but i can't get myself clear why the two events mentioned are mutually exclusive.
I thought the following events are mutually exclusive for this problem, and i arrived with the following solution
1) 1 person from 2-sibling group and 1 person from 1-sibling group, which is the same with your second event:
we have 3/7 probability of getting 2-siblings person and 4/6 probability of not getting his or her sibling.
2) No one from the 2-sibling group.
we have 4/7 probability of getting 1-sibling person and 2/6 probability of not getting his or her sibling and not within one of 2-sibling group.
p = 3/7*4/6 + 4/7*2/6 = 20/42 (this is not within the answers, so i guess it is wrong, but could you please help me where I got wrong with this). Many thanks.

3) 1 person from 1-sibling group and next person from 2-sibling group: 4/7*3/6

p = 20/42 + 12/42 = 16/21

Actually, you have three combinations: 1) 21 2) 11 3) 12.
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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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05 Jul 2010, 14:52
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5
Four people have exactly one sibling and three people have exactly two siblings.

Think about how this can be done...the only way to organize is this is to have two pairs of siblings and one trio of siblings. In other words:

AB (one sibling pair)
CD (another sibling pair)
EFG (a trio of siblings)

This way each of A and B are siblings, each of C and D are siblings, so we have four people each with exactly one sibling. And each of E, F, and G are siblings so we have three people each of whom has exactly two siblings.

Because this is a "NOT" probability question, we should see how many ways we CAN select two individuals who are siblings.

Well, we can select the AB pair or the CD pair. So far, that's 2 ways. We can also select any two people from the EFG trio, so that's another 3C2 or 3 ways.

So, there are a total of 2+3 = 5 ways of pulling out 2 individuals who are siblings.

Probability = (#desired)/(#total)

The denominator of the formula is just all the ways we can select any two people from the seven. So we have:

Probability of selecting 2 people who are siblings = 5/7C2 = 5/21

Therefore, the probability of selecting 2 people who are NOT siblings is:

1 - 5/21 = 16/21

Choose 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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05 Jul 2010, 17:42
2
That is one way, I think it is easier to calculate the probabilities than the combinations.

there is 4/7 chance of selecting a person with one sibling initially with a 5/6 probability the second person will not be a sibling.

There is a 3/7 Chance of selecting a person with 2 siblings initially with 4/6 probability the second person will not be a sibling.

4/7 * 5/6 + 3/7 * 4/6 = 20/42 + 12/42 = 32/42 = 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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11 Aug 2010, 00:06
2
Jinglander wrote:
In a room filled with 7 people, 4 have 1 sibling and 3 have two siblings. If two people at selected at random what is the prob that they are not siblings.

Can someone explain.

Out of 7 people
4 have 1 sibling So these 4 form 2 pairs of siblings
3 have 2 siblings So all these 3 are siblings to each other.

So the 7 people are like A1A2, B1B2, C1C2C3 where A1 and A2 are siblings to each other , B1 and B2 are siblings to each other and C1,C2 and C3 are siblings to each other.

Probability of selecting 2 people who are not siblings
= 1 - Probability of selecting 2 people who are siblings
= 1 - $$(C^2_1+C^3_2)/C^7_2$$ (Select either A1A2/B1B2 or any 2 out of C1C2C3)
= 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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11 Aug 2010, 01:59
2
total possibilities 2C7=21
possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5
probability 1-5/21
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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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13 Sep 2010, 19:36
total possibilities 2C7=21
possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5
probability 1-5/21

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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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13 Sep 2010, 19:57
4
3
harithakishore wrote:
total possibilities 2C7=21
possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5
probability 1-5/21

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).

Let's calculate the probability of opposite event and subtract it from 1. Opposite event would be that chosen 2 individuals are siblings.

# of selections of 2 out of 7 - $$C^2_7=21$$;
# of selections of 2 siblings - $$C^2_3+C^2_2+C^2_2=3+1+1=5$$, here $$C^2_3$$ is the # of ways to choose 2 siblings out of siblings 5-6-7, $${C^2_2}$$ is the # of ways to choose 2 siblings out of siblings 1-2, and $$C^2_2$$ is the # of ways to choose 2 siblings out of siblings 3-4;

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

You can check other approaches in my first post.

Hope it's clear.
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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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01 Apr 2012, 01:16
1
u have 2 groups
(a b x y) and (c d e)

note, that a=b (siblings)
c=d=e (siblings)

to find no two identical siblings, u need to find 2 no siblings in group 1 , or 1 sibling from group 1 and 1 sibling from group 2

1.to find 2 no siblings in group 1 (a b x y)-

u have 4 such combinations - (a;x) (a;y) (b;x) (b;y)

2. 1 sibling from group 1 and 1 sibling from group 2 -

4C1*3C1=4*3=12

(4+12)/7C2=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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13 Jun 2012, 23:25
Bunuel wrote:
alchemist009 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

Bunuel,

can you explain your #3 approach.

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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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14 Jun 2012, 00:28
1
1
Rice wrote:
Bunuel wrote:
alchemist009 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

Bunuel,

can you explain your #3 approach.

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.
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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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30 Jun 2012, 20:51
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) $$=4+6+6=16$$.

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

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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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01 Jul 2012, 03:15
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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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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12 Nov 2012, 05:38
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1
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 sibling in the  [#permalink]

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13 Dec 2012, 15:08
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 sibling in the  [#permalink]

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15 May 2013, 01:26
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
Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the   [#permalink] 15 May 2013, 01:26

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