probability

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dbag

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10 people are standing a double line. what is the probability that 2 random people are next to each other or in front or behind of the other?

probability for next to each other is: (9!*2!)/10! so 0.2

what about in front or behind? how do i figure that out?

total probability is then 0.2 + the probability for case 2
 
dbag said:
10 people are standing a double line. what is the probability that 2 random people are next to each other or in front or behind of the other?

probability for next to each other is: (9!*2!)/10! so 0.2

what about in front or behind? how do i figure that out?

total probability is then 0.2 + the probability for case 2
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Could you explain your calculations?
 
lev888 said:
Could you explain your calculations?
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10! ways to arrange people, 9! for the pair that stands next to each other and 2! since they can stand in 2 ways. therefore 1/5 change that 2 random people stand next to each other
 
dbag said:
10! ways to arrange people, 9! for the pair that stands next to each other and 2! since they can stand in 2 ways. therefore 1/5 change that 2 random people stand next to each other
Click to expand...
How do you get 9!?
 
One is going to have to make an assumption about the structure of the lines. Are they both 5 people long?
 
5 slots for being next to one another. 2 ways to arrange the 2 people per slot. 10 ways.

4 slots for being front/back of one another. 2 lines, 2 ways to arrange w/in a slot. 16 ways.

26 total ways to be adjacent between the lines or within a line.

90 total possible arrangements of the two people. (pick one of 10 queue positions, then pick one of 9)

\(\displaystyle p = \dfrac{26}{90}=\dfrac{13}{45}\)
 
i see. i was doing the first part wrong aswell. thanks very much
 
dbag said:
i see. i was doing the first part wrong aswell. thanks very much
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A simpler approach for the first part - given any position of person A, how many positions for B are next to A?
 
lev888 said:
A simpler approach for the first part - given any position of person A, how many positions for B are next to A?
Click to expand...

You need to be a little careful though. 4 spots have different numbers of adjacent positions than the rest.
 
Romsek said:
the fronts and backs of the lines have only two spots that are adjacent to them.

the rest of the spots have 3
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I agree. But I am referring to part 1 only and assuming that next is side by side.
 
Romsek said:
5 slots for being next to one another. 2 ways to arrange the 2 people per slot. 10 ways.

4 slots for being front/back of one another. 2 lines, 2 ways to arrange w/in a slot. 16 ways.

26 total ways to be adjacent between the lines or within a line.

90 total possible arrangements of the two people. (pick one of 10 queue positions, then pick one of 9)

\(\displaystyle p = \dfrac{26}{90}=\dfrac{13}{45}\)
Click to expand...
Nicely done.
 
I think the answer is as follow:
P1=Probability of being next to each other=1577137804739.png=8/45
P2=Probability of being in front of each other=1577137857319.png=5/45
total answer=P1+P2=1577137956989.png
 
Your numbers are correct, but in the wrong places. Can you explain the reasoning behind your calculations?

I myself solved it by the method in post #9 (except that I used combinations rather than permutations). If I approached it your way, I would use very different calculations for each part. The probability that they are next to one another is just 1/9: wherever the first one is, there is one place, out of 9, where the other would be next to them. The probability that one is in front of the other is a little harder.
 
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