Probability

[heshamelaziry]

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 [spoiler]16/21[/spoiler]

[Abdulla]

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 [spoiler]16/21[/spoiler]

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 #

[heshamelaziry]

HOhw you say you are bad with probability ? you got the right answer. However, I am still lost.

[Abdulla]

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.

[heshamelaziry]

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.

Still very well done. I still can't rationalize it . BTW, did u get my pm ?

[getso]

Hi ,

Could you please explain the logic...

[viju9162]

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

[mridul_dave]

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 [spoiler]16/21[/spoiler]

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.

[getso]

Thanks for the great explaination ..

[Abdulla]

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 [spoiler]16/21[/spoiler]

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