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Please help Probability!!

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by [email protected] » Wed Aug 06, 2008 8:39 am
In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

Thanks in advance!

A)5/21
B)3/7
C)4/7
D)5/7
E)16/21

OA A
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by geoshieh » Wed Aug 06, 2008 9:58 am
There are 7 total ppl and we are trying to find to total number of combinations that they could make so 7 combination 2 = 21

Then count out the total number of successful pairs which is 5 so that gives 5/21 successful pairs, and 1 - 5/21, which is 6/21 unsuccessful pairs

E)16/21


You can also use the binomial coefficient to solve this.
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by 2008 » Wed Aug 06, 2008 10:05 am
mmmm i did it in a different way but cannot find a way out... can someone help?

let's say that the one who has one friend are: A B C D
let's say that the one who has two friends are: E F G

hence
E is friend with A & B
F is friend with C & D
G is friend with F & E

now the probability of having two of the first group is: 4/7*3/6= 2/7

obviously wrong... where is my mistake?
thanks
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by parallel_chase » Wed Aug 06, 2008 10:15 am
2008 wrote:mmmm i did it in a different way but cannot find a way out... can someone help?

let's say that the one who has one friend are: A B C D
let's say that the one who has two friends are: E F G

hence
E is friend with A & B
F is friend with C & D
G is friend with F & E

now the probability of having two of the first group is: 4/7*3/6= 2/7

obviously wrong... where is my mistake?
thanks
If you look closely E has A, B, & G as friends, which is not possible because any 3 people can only have max 2 friends.
Similarly F has C, D, E & G as friends, which is also not possible.

Here is a way to look at this

A - B
C - B

A has one friend B
C has one friend B
B has two friends A & C

E-F
G-F

E has one friend F
G has one friend F
F has two friends E & G

D-A
D had one friend A
A has two friends B & D

Now we have 4 people (C,D,E,G) with 1 friend each and 3 people (B,A,F) with 2 friends each.

We have to find the probability of picking C and D or E & G who are not friends with each other.

If you start calculating that it will be really lengthy and tedious task, therefore find the probability of picking two friends together and subtract that by 1.

Hope this helps.
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by [email protected] » Thu Aug 07, 2008 2:29 am
parallel_chase thanks for the response.

Please can elaborate on how to go about findign the probability of picking two friends together and subtract that by 1.

Thanks
Karen
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by parallel_chase » Thu Aug 07, 2008 6:34 am
[email protected] wrote:parallel_chase thanks for the response.

Please can elaborate on how to go about findign the probability of picking two friends together and subtract that by 1.

Thanks
Karen
We have 4 people (C,D,E,G) with 1 friend each and 3 people (B,A,F) with 2 friends each.

If you look at above case the letters will shuffle but we will always have 4 letters with 1 friend each and 3 people with 2 friends each.

Case I
Probability of picking 2 friends from 4 people who have 1 friend each.

Probability of picking 1st person from 4 = 4/7
Probability of picking his friend = 1/6
(4/7)*(1/6) = 2/21

Probability of picking 2 friends from 3 people who have 2 friend each.

Probability of picking 1st person from 3 = 3/7
Probability of picking his friend = 2/6

(3/7)*(2/6)=1/7

Total probability = (2/21) + (1/7) = 5/21

1-(5/21) = 16/21


Hope this helps. Let me know if you have any doubt.
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by sudhir3127 » Thu Aug 07, 2008 7:00 am
This is how i approached it...

We have a 4/7 chance of picking people who have only 1 friend.
Now there is a 5/6 chance of picking the second person who's not their friend (6 people , 1 is friend, thus 5 are not).

We have a 3/7 chance of picking people who have 2 friends.
4/6 chance of picking a second person who's not their friend.

(4/7)(5/6)+(3/7)(4/6)=32/42=16/21

Thus E..

do let us know if u have any doubts...
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by ildude02 » Thu Aug 07, 2008 12:50 pm
geoshieh wrote:There are 7 total ppl and we are trying to find to total number of combinations that they could make so 7 combination 2 = 21

Then count out the total number of successful pairs which is 5 so that gives 5/21 successful pairs, and 1 - 5/21, which is 6/21 unsuccessful pairs

E)16/21


You can also use the binomial coefficient to solve this.
How did you get 5 as as the successfull pairs using combinations? Appreciate your response.
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by eccentric » Fri Aug 08, 2008 9:53 am
Let say A,B,C,D have one friend each and E,F,G two friends each
let us define friendship now
{AB}{CG}{DE}{EF}{FG}

two people can be chosen from 7 in 7C2 ways:-> 21
so probabitliy that the two chosen are not freinds = 1- p{being friend}
1-5/21 = 16/21

Hope this helps!!!!!
but i fail to understand why OA is A
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