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

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 - \(C^2_3+C^2_2+C^2_2=3+1+1=5\);

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

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.

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.

@ 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. _________________

Re: Sibling Problem [#permalink]
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 _________________

___________________________________ Please give me kudos if you like my post

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;

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

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: PS-Probability [#permalink]
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.

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.

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}.

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}.

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

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.

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.

Re: In a room filled with 7 people, 4 people [#permalink]
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?

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

gmatclubot

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

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