Probability

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heshamelaziry
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Topic: Probability
PostMon Nov 16, 2009 10:58 pm

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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?

OA 16/21
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PostTue Nov 17, 2009 12:12 am

heshamelaziry 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?

OA 16/21
I am bad with Probabilities but I will try it..

Prob = # desire outcomes / Total outcomes

Let's start with the easy one. We want to choose 2 out of 7.

Total outcomes = 7! / 2!5! = 21 ( 7 x 6 x 5 x 4 x 3 x 2 x 1 / 5 x 4 x 3 x 2 x 1 x 2 x 1 )

# of Desired outcomes. We've got two scenarios.
1) Choosing two people from the first group.
4! / 2!2! = 6 ..... ( remember that we've selected 2 people each of which has one sibling in the room, so we should subtract 2)
6-2 = 4 .......(1)

2) Choosing one from the first group times another from the second group.
4! /1!3! * 3! /2!1! = 4*3 = 12 ...........(2)

4+12 / 21 = 16/21 #

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heshamelaziry
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PostTue Nov 17, 2009 12:34 am

HOhw you say you are bad with probability ? you got the right answer. However, I am still lost.
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Abdulla
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PostTue Nov 17, 2009 12:49 am

heshamelaziry wrote:
HOhw you say you are bad with probability ? you got the right answer. However, I am still lost.
I spent 15 minutes to figure it out. Very Happy
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heshamelaziry
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PostTue Nov 17, 2009 12:55 am

Abdulla wrote:
heshamelaziry wrote:
HOhw you say you are bad with probability ? you got the right answer. However, I am still lost.
I spent 15 minutes to figure it out. Very Happy
Still very well done. I still can't rationalize it Mad . BTW, did u get my pm ?
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PostTue Nov 17, 2009 5:19 am

Hi ,

Could you please explain the logic...
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PostTue Nov 17, 2009 5:32 am

Hi Abdulla, heshamelaziry,

I didnt understand how you arrived at the answer. Can you explain how can we solve with this formula:

P (A) = 1 - P`(A).

Regards,
Viju

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PostTue Nov 17, 2009 7:35 am

heshamelaziry 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?

OA 16/21
Lets say letters represents a person and a '--' represents the sibling relation.

A -- B
B -- C
these are the 4 people who have exactly one sibling.


E -- F
\ /
G

There are the three people with exactly two siblings each. We have total 7 people.

We just made three sets of people.

Lets get the prob. that two are siblings. For that either you pick from the first set, or second set or any two from the third set.

2C2 + 2 C2 + 3C2 = 1 + 1 + 3 = 5

Total ways is 7C2 = 21. Prob that two are siblings is 5/21.
So prob they are not is 1 - 5/21 = 16/21.
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PostTue Nov 17, 2009 7:51 am

Thanks for the great explaination ..

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Abdulla
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PostWed Nov 18, 2009 12:29 am

mridul_dave wrote:
heshamelaziry 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?

OA 16/21
Lets say letters represents a person and a '--' represents the sibling relation.

A -- B
B -- C
these are the 4 people who have exactly one sibling.


E -- F
\ /
G

There are the three people with exactly two siblings each. We have total 7 people.

We just made three sets of people.

Lets get the prob. that two are siblings. For that either you pick from the first set, or second set or any two from the third set.

2C2 + 2 C2 + 3C2 = 1 + 1 + 3 = 5

Total ways is 7C2 = 21. Prob that two are siblings is 5/21.
So prob they are not is 1 - 5/21 = 16/21.
WOW, Easy approach..
Thanks hesham

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