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Probability

by nikhilkatira » Tue Jun 08, 2010 3:17 am
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
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by liferocks » Tue Jun 08, 2010 3:43 am
in the room 2 pairs of sibling and 3 siblings are present
if 2 individuals so that they are siblings can be done by selecting any one of the 2 pair in 2 ways any any 2 of the 3 siblings in 3C2=3 ways
Total 5 ways

2 people can be selected out of 7 in 7C2=21 ways

so If two individuals are selected from the room at random, the probability that those two individuals are NOT siblings is
= 1- 5/21 or 16/21

IMO ans option E
"If you don't know where you are going, any road will get you there."
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by jube » Tue Jun 08, 2010 4:04 am
Probability that the 2 people will be siblings:

1) If it's from amongst the 4 people with only 1 sibling each:

(1/7)(1/6)4 = 2/21

2) If it's from amongst 3 people with 2 siblings each:

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

Total Probability that 2 people will be siblings = 2/21+1/7=5/21
Probability that 2 people are not siblings = 1- (5/21) =16/21

E
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by Thouraya » Wed Jun 09, 2010 1:18 pm
Testluv,

In the case of option 1 where the 2 siblings are chosen from the 4 people, how did u calculate it to be equal 2 ways? Its a combination but Im not sure how to get it...At first I put it, 4C2; thus 6 ways but its wrong I know..the 4 individuals each have a sibling, and having to choose 2 siblings is what confused me..


Thanks!
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by Testluv » Wed Jun 09, 2010 11:11 pm
Thouraya wrote:Testluv,

In the case of option 1 where the 2 siblings are chosen from the 4 people, how did u calculate it to be equal 2 ways? Its a combination but Im not sure how to get it...At first I put it, 4C2; thus 6 ways but its wrong I know..the 4 individuals each have a sibling, and having to choose 2 siblings is what confused me..


Thanks!
Hi!

Well, I didn't actually post in this thread!

But 4 people each have exactly 1 sibling while 3 people each have exactly 2 siblings. In loose combinatorics/probability questions, I often like to use slots to visualize the situation:

___..___............___..___.................___..___..___

In "NOT" and "at least" combinatorics/probability questions, it is often helpful to think of the converse. So, how many ways CAN we have 2 siblings? Well, we can select the first sibling pair OR the second sibling pair OR any 2 from the sibling trio. That's 2C2 + 2C2 + 3C2, or more simply 1 + 1 + 3 ways, for a total of 5 ways we can select 2 siblings (as liferocks points out).

Prob = #desired/#total

There are 7C2 or 21 total ways of selecting any 2 objects from 7.

So the probability of selecting a pair of siblings is 5/21. Thus, the probability of NOT selecting a pair of siblings is 16/21.

Let me know if it's not clear.
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by Thouraya » Sat Jun 12, 2010 10:58 am
I got the approach, and the 3C2 and the 7C2...but what I am finding it hard to get is the rationale regarding the 4 people and why 2C2

Lets assume these are the four people:

A B C D

Because each has one relative, lets assume A is relative B just like B is relative A, and C is relative D just like D is relative C, correct?

So whether A is relative B or B is relative A, its the same, why are we assumig theyre two different options?

Thank you:)
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by Testluv » Sat Jun 12, 2010 12:28 pm
Thouraya wrote:I got the approach, and the 3C2 and the 7C2...but what I am finding it hard to get is the rationale regarding the 4 people and why 2C2

Lets assume these are the four people:

A B C D

Because each has one relative, lets assume A is relative B just like B is relative A, and C is relative D just like D is relative C, correct?

So whether A is relative B or B is relative A, its the same, why are we assumig theyre two different options?

Thank you:)
We're not assuming that AB and BA are different pairs! They are the same pair. AB is one sibling pair and CD is another. That's two sibling pairs. And if E, F, and G are all siblings, then we can have EF, EG or FG--that's another three sibling pairs for a total of 5 possible sibling pairs.

Or: In how many ways can we choose 2 siblings? Well we can choose 2 from the AB sibling pair--that's 2C2 or 1. We can choose 2 from the CD sibling pair--that's 2C2 or 1. Or we can choose any 2 from the trio of siblings (E, F, G)--that's 3C2 or another 3. For a total of 5 ways.

______


Regarding your earlier post, we would use 4C2 if we had a quartet of individuals all of whom were siblings of one another; ie, A,B,C, and D are all siblings.
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by kvcpk » Thu Jun 17, 2010 7:17 am
liferocks wrote:in the room 2 pairs of sibling and 3 siblings are present
if 2 individuals so that they are siblings can be done by selecting any one of the 2 pair in 2 ways any any 2 of the 3 siblings in 3C2=3 ways
Total 5 ways

2 people can be selected out of 7 in 7C2=21 ways

so If two individuals are selected from the room at random, the probability that those two individuals are NOT siblings is
= 1- 5/21 or 16/21

IMO ans option E
How does "4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room"
convert to
"in the room 2 pairs of sibling and 3 siblings are present" ?
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by ssuarezo » Thu Jun 17, 2010 11:41 am
kvcpk wrote:
liferocks wrote:in the room 2 pairs of sibling and 3 siblings are present
if 2 individuals so that they are siblings can be done by selecting any one of the 2 pair in 2 ways any any 2 of the 3 siblings in 3C2=3 ways
How does "4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room"
convert to
"in the room 2 pairs of sibling and 3 siblings are present" ?
hI kvcpk:

"4 people have exactly 1 sibling in the room " means

A B C D people

each has one siblig (A,B), (B,A), (C,D) (D,C), which means "2 pairs of siblings"

"and 3 people have exactly 2 siblings in the room"

A B C

If A has B, C --> B has AC --> C has AB, which means " 3 siblings are present "

Silvia.
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by kvcpk » Thu Jun 17, 2010 11:52 am
Got it now.. Thanks!!
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