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

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

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01 Nov 2009, 09:23
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Feb 2012, 21:48, edited 1 time in total.
Edited the question and added the OA
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01 Nov 2009, 09:51
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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: MGMAT CAT - Picking Friends [#permalink]

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02 Dec 2009, 20:27
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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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01 Nov 2009, 18:28
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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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13 Sep 2010, 18:57
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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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05 Jul 2010, 13:52
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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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10 Aug 2010, 23:06
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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 [#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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13 Jun 2012, 23:28
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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 [#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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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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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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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: MGMAT CAT - Picking Friends [#permalink]

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02 Dec 2009, 21:50
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: MGMAT CAT - Picking Friends [#permalink]

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03 Dec 2009, 04:58
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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05 Jul 2010, 16:42
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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05 Jul 2010, 17:39
Great explanations guys.

Posted from my mobile device
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11 Aug 2010, 00:59
total possibilities 2C7=21
possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5
probability 1-5/21
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13 Sep 2010, 18:36
total possibilities 2C7=21
possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5
probability 1-5/21

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13 Sep 2010, 18:58
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

There are suppose A B C D E F G members in the room 4 people who have exactly one sibling....A B C D....(A is Bs sibling and viceversa) (C is Ds sibling and viceversa)...now remaning EFG are 3 people who have exactly 2 siblings....(E has F and G as his/her sibling and so on..)
there are now 3 different set of siblings (A and B)
(C and D);(EFG)
Now first selecting 2 people out of 7 is 7C2=21
first sibling pair----(A and B)--selecting 2 people --2C2=1
second sibling pair (C and D)--selecting 2 people--2C2=1
third sibling pair (E F G)--selecting 2 out of 3 --3C2=3

total= 1+1+3=5
but,a/c to formula P(success)-1-p(fail)
here,p(failure)is selecting 2 people who are siblings
=5/21(21 is 7C2)
=1-5/21
=16/21
ANS E
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16 Dec 2010, 05:54
first, find probability of getting two pairs and triplet  4/7*1/6=4/42 and 3/7*2/6=6/42
second, add these two results  4/42 + 6/42 = 10/42 = 5/21
third, subtract the result of addition from 1  1 – 5/21 = 16/21
Re: Probability   [#permalink] 16 Dec 2010, 05:54

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