Question

Need Some Help, Number Line Probability:

A point is chosen at random on the number line between 0 and 1, and the point is colored green. Then another point is chosen at random on the number line between 0 and 1, and this point is colored purple, what is the probability that the number of the purple point is greater than the number of the green point, but less than twice the number of the green point?

Answer 1

good one... what is the distribution of the green point?

Answer 2

I'm not quite sure what you are asking

Answer 3

you need to think about what the underlying distribution (probability distribution) of the green point is in order to get a handle on the probability you seek.

Answer 4

have you heard of the uniform distribution?

Answer 5

also, have you heard of conditional probabilities?

Answer 6

No, I haven't, I might have learned it, just not under that name.

Answer 7

is this a calculus based course?

Answer 8

No

Answer 9

Although my course does provide a solution after you correctly answer the problem.

Answer 10

\[X \text{ ~ } U \left( a, b \right) => P \left( X<x \right)=\frac{ 1 }{ x-a } \text{ for } a<x<b\]
this is the continuous version of the uniform distribution which is what you'd need to do this probability. You don't necessarily need calculus to do it but you do need to know about conditional probability.


It's difficult to write in this equation editor... let me see if I can work something up for you.

Answer 11

Hmm, I'm beginning to understand that part. For a, we could put the green point, x, the purple point, and b 2 times the green point... unless I am missing something.

Answer 12

you're getting there...

are the events independent?

Answer 13

Hmm, I believe so, I'm not sure though

Answer 14

yes, they are. why, beacuse no restrictions are imposed once we pick our green point. I mean, the purple point can still be anywhere from 0 to 1.
that means P( G and P) = P(G) * P(P)
but we should still condition on what we get for green.
if green is less than .5 then we're looking for
P( g<P<2g)
and if green is >= .5 then we want
P(P>g) = 1-P(G<g)

Answer 15

What is the difference between G and g? (same with P and p, but not P as in probability?

Otherwise, I understand it

Answer 16

@pgpilot326 If the events are independent then the probability is not conditional....

Answer 17

Unless you're referring to the fact that if the green point is chosen on a higher number, then the probability of the purple point being greater automatically reduces thus conditional...

Answer 18

the caps are the random variable
the smalls are the values they take on

yes, right @genius12 but if we condition, it's much easier to get a handle on what's needed

Answer 19

in that first one
P(g<P<2g), g represents that we got an actual number for G. For example, let's say g = .33, then we can compute P(g<P<2g) = P(.33<P<.66) = (.66 - .33)

The distribution I gave earlier is erroneous and I apologize.

It should be the length of the interval you want over the total length of the interval.

In this problem the total interval is 1. I apologize.

Answer 20

Ah, okay.

Answer 21

Hmm, I think i've found a way, using graphing

Answer 22

Because if x = green point, y = purple point

Answer 23

0 <= x <= 1, 0 <= y <= 1

Answer 24

Then, you would graph the inequality x <= y <= 2x