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In a room filled with 7 people, 4 people have exactly 1 [#permalink]
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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
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Originally posted by reply2spg on 01 Nov 2009, 10:23.
Last edited by Bunuel on 14 Feb 2012, 22:48, edited 1 time in total.
Edited the question and added the OA



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In a room filled with 7 people, 4 people have exactly 1 [#permalink]
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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 (12; 34). As there are 3 people with exactly 2 siblings each: we have one triple of siblings (567). 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]
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01 Nov 2009, 19:28
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You have people 1234567
4 have exactly 1 sibling which can mean: 12 are siblings 34 are siblings
3 have exactly 2 siblings which can mean: 567 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 1234 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 nonsibling pair is 1234 Therefore the probability is 4/42. Multiply that probability by 3, which represent 567 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]
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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 1sibling person and 5/6 probability of not getting his or her sibling. 2) we have 3/7 probability of getting 2siblings 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]
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02 Dec 2009, 22:50
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 : 15/1 =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 2sibling group and 1 person from 1sibling group, which is the same with your second event: we have 3/7 probability of getting 2siblings person and 4/6 probability of not getting his or her sibling. 2) No one from the 2sibling group. we have 4/7 probability of getting 1sibling person and 2/6 probability of not getting his or her sibling and not within one of 2sibling 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, 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 2sibling group and 1 person from 1sibling group, which is the same with your second event: we have 3/7 probability of getting 2siblings person and 4/6 probability of not getting his or her sibling. 2) No one from the 2sibling group. we have 4/7 probability of getting 1sibling person and 2/6 probability of not getting his or her sibling and not within one of 2sibling 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 1sibling group and next person from 2sibling 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: PS probability 700800 [#permalink]
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05 Jul 2010, 14:52
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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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Re: PS probability 700800 [#permalink]
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05 Jul 2010, 17:42
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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.
Simply adding these probabilities together gives you the answer.
4/7 * 5/6 + 3/7 * 4/6 = 20/42 + 12/42 = 32/42 = 16/21



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Re: Sibling Problem [#permalink]
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11 Aug 2010, 00:06
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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]
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11 Aug 2010, 01:59
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total possibilities 2C7=21 possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5 probability 15/21



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Re: Probability [#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 15/21
vittar..can you please explain 2C3.....



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Re: Probability [#permalink]
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harithakishore wrote: total possibilities 2C7=21 possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5 probability 15/21
vittar..can you please explain 2C3..... As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (12; 34). As there are 3 people with exactly 2 siblings each: we have one triple of siblings (567). 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 567, \({C^2_2}\) is the # of ways to choose 2 siblings out of siblings 12, and \(C^2_2\) is the # of ways to choose 2 siblings out of siblings 34; \(P=1\frac{5}{21}=\frac{16}{21}\). You can check other approaches in my first post. Hope it's clear.
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Re: Probability [#permalink]
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13 Sep 2010, 19:58
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 people2C2=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)1p(fail) here,p(failure)is selecting 2 people who are siblings =5/21(21 is 7C2) =15/21 =16/21 ANS E



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Re: Probability [#permalink]
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16 Dec 2010, 06:54
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: PSProbability [#permalink]
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14 Aug 2011, 17:04
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]
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01 Apr 2012, 01:16
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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]
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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
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]
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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: inaroomfilledwith7people4peoplehaveexactly87550.html#p645861Hope it's clear.
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Re: Probability [#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 1i.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 2Similarly, 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 3Last 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 (12; 34). As there are 3 people with exactly 2 siblings each: we have one triple of siblings (456).
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]
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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 1i.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 2Similarly, 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 3Last 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 (12; 34). As there are 3 people with exactly 2 siblings each: we have one triple of siblings (456).
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 doublecount some cases. Hope it's clear.
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