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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] New post 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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Re: Probability [#permalink] New post 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}.

Answer: E.
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Re: Probability [#permalink] New post 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

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
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Re: MGMAT CAT - Picking Friends [#permalink] New post 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

maybe this will be helpful:

- Probability
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Re: MGMAT CAT - Picking Friends [#permalink] New post 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] New post 03 Dec 2009, 04:58
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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.


In your approach you need to add one more event:

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: Sibling Problem [#permalink] New post 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.

Answer is 6/21

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: Sibling Problem [#permalink] New post 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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Re: Probability [#permalink] New post 13 Sep 2010, 18:36
total possibilities 2C7=21
possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5
probability 1-5/21

vittar..can you please explain 2C3.....
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Re: Probability [#permalink] New post 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

vittar..can you please explain 2C3.....


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: Probability [#permalink] New post 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 B`s sibling and viceversa) (C is D`s 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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Re: Probability [#permalink] New post 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
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Re: PS-Probability [#permalink] New post 14 Aug 2011, 16:04
there are 7 people in the room.

there are 4 ppl with exactly one sibling. lets consider the following
1,3 - group1
2,4 - group 2
there are 3 people with exactly two siblings
5,6,7 - group 3

so the probability of picking 2 non siblings is nothing but picking each from different groups.

= (2c1*2c1 + 2c1*3c1 +3c1*2c1)/7c2 = 16/21

Answer is E.
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Re: Toughest MGMAT Math Problems (2/20) [#permalink] New post 01 Apr 2012, 00:16
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

hope it helps,but if u have any question, please, ask.
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Re: sibling [#permalink] New post 13 Jun 2012, 22:25
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

how do i figure it out in simplest manner? please help


Merging similar topics. Please ask if anything remains unclear.



Bunuel,

can you explain your #3 approach.

Thanks in advance.
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Re: sibling [#permalink] New post 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

how do i figure it out in simplest manner? please help


Merging similar topics. Please ask if anything remains unclear.



Bunuel,

can you explain your #3 approach.

Thanks in advance.


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: Probability [#permalink] New post 30 Jun 2012, 19:51
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}

Answer: E.

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Re: Probability [#permalink] New post 01 Jul 2012, 02:15
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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}

Answer: E.


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: Probability [#permalink] New post 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] New post 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
Re: In a room filled with 7 people, 4 people   [#permalink] 03 Nov 2012, 16:36
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