- Kainui

Here's a fun problem I came across. During an hour two independent events can happen at any time. What's the probability that the events are at least 10 minutes apart?

- jamiebookeater

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- rational

|dw:1432999492312:dw|

- Kainui

Hahaha yeah you got it.

- rational

P(|E1-E2| > 10 minutes) = (Area of shaded region) / (Total area) = 5^2/6^2 = 25/36 ?

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## More answers

- Kainui

Yep, exactly. =)

- rational

these are really fun xD
i got some practice a couple of months ago when @amistre64 was preparing for some probability theory exam

- Kainui

Something that just occurred to me that I don't know the answer to is what is the probability of having 3 events 10 minutes apart?
Or, is there a good reason why the probability is 5^2/6^2? If the events were separated by 20 minutes would the answer then become 4^2/6^2 ? I guess this would be easy to check for the general answer.

- rational

very interesting, 3 events requires a triple integral is it

- rational

second part of the question is easy yeah
we always get squares in top and bottom because of symmetry : |E1-E2| > 20

- Kainui

for the 2D case in general the answer is:
\[\frac{(60-t)^2}{60^2}\] which is quite nice!

- ParthKohli

Wow, that's beautiful.

- ParthKohli

Looks like my book has this... is this what is called "infinitistic probability"?

- Kainui

Yup, I guess? I guess continuous probability distribution maybe? It's important for quantum mechanics!

- ikram002p

:)

- dan815

|dw:1433020038430:dw|

- dan815

oh greater so 5/6

- dan815

|dw:1433020148264:dw|

- dan815

is that it?

- dan815

i see that for one case its, 5/6 then therefore for every other point its always 5/6 of the total there so 5/6 all the time :)

- dan815

i solved it by looking at a number line

- dan815

|dw:1433020285038:dw|

- dan815

so i thought about summing all these individual events up (kind of like an integral) however its constant here its always 5/6 of the smaller case, that event 1 happens at the one single point

- dan815

Hence 5/6 for the complete thing

- dan815

here is also something interesting... prolly not so much but kinda cool to visualze lol

- dan815

cause its not good to think ofa number line but a circular number like for this or like a clock

- dan815

|dw:1433020557004:dw|

- dan815

|dw:1433020604712:dw|

- Kainui

the problem is your answer seems to be wrong, the answer is 25/36 not 5/6 for the first question

- dan815

ya i got that i change it xD

- dan815

for a sec i thought u were asking less than 10

- dan815

oh dang really

- dan815

why is it 5^/6^2?

- dan815

lemmee think

- Kainui

this picture: |dw:1433020730099:dw| imagine each point is when two events happen, so all the points on the diagonal happen simultaneously. so for instance when one event happens, (10,20) at that point, that will fall in the black, so it's greater or equal to 10 min right?
So the ratio of the black region to the entire square is the probability.

- dan815

oh :O i get it

- dan815

theres a problem here with assuming the circular thing too, because once event 1 occurs later

- dan815

the prob is greater for 10 mins

- dan815

:D

- dan815

or wait no, the even can happen back in time too

- Kainui

This is pretty similar to the Buffon needle problem which is a cool way to approximate \(\pi\).

- dan815

i dont think i understand the graph

- Kainui

Honestly I don't think it gets much better than this general case:
\[\frac{(60-t)^2}{60^2}\]
this tells you the probability that two random events happen with at least a separation of time t within an hour.
Like if t=10 in the original then we get: \[\frac{(60-10)^2}{60^2} = \frac{25}{36}\] and of course the probability that they're 0 minutes apart will give us 100% and probability they are 60 min apart will give us 0% just like we expect, it's just from calculating it based off of a variable triangle size inside cut out basically lol.

- Kainui

|dw:1433021207310:dw| Here are some random points picked in there. The first represents some time that one event happened and the other represents the time the second event happened.
So look along the diagonal, (0,0) means both events happened at hte beginning of the our. (30,30) means both events happened at the 30 minute mark. Makes sense?
Now look at the point (10,50) that means the first event happened at 10 minutes and the other event happened at 50 minutes. This is one of the points that falls within "both events happening 10 minutes apart or greater" since these are obviously separated by 40 minutes yeah?

- dan815

ah i gotcha now xD

- dan815

the graph labelling didnt quite make sense to me, now its clear thanks

- Kainui

cool so what if we have 3 events separated by 10 minutes, what does that cube look like?

- dan815

|dw:1433021555908:dw|

- Kainui

You would think so, that was my first guess too, but it doesn't seem to be right if you think of evaluating say, this point: |dw:1433021616607:dw|

- dan815

ok lemme see more algrebrically first
a+b+c<60*3
and
diff between a,b,c is greater than 10

- dan815

(5/6)^3?

- dan815

^ thats a guess

- dan815

wait what is the question?is it all 3 have to be apart by 10? or as long as all 3 dont happen in 10

- Kainui

|dw:1433021690423:dw|

- dan815

ok wat u drew there looks like its as long all 3 dont happen within 10, so u can have 2 with in 10 and other out

- Kainui

So one example of an event can be like (10, 20, 30) since they are all separated by 10 min.

- dan815

|dw:1433021882772:dw|

- Kainui

yeah what I drew is wrong I think idk haha

- Kainui

Yeah I just didn't wanna draw it with those lines in there cause it looked tacky

- dan815

|dw:1433021915541:dw|

- dan815

lol

- dan815

oh u can think of the 2by 2 case as 1 big event again

- dan815

|dw:1433022504943:dw|

- dan815

so u solve for cases, when event 1 and 2 are close, and when they are not

- dan815

eh nvm i feel like there is something easier to compute off the differences

- dan815

did u already solve it?

- Kainui

Nahhhh

- dan815

ok :) im on it then sir

- Kainui

some useful things:
\( 20 \le x+y+z \le 40\)
\( 10 \le x+y \le 50\)
\( 10 \le x+z \le 50\)
\( 10 \le y+z \le 50\)

- dan815

that what i was lookign at

- dan815

like can we write out system of equationa for lines

- dan815

and break it down into 4 or 6 or 9 cases or something

- Kainui

I think this is enough to make up an integral, those are our bounds

- dan815

hey btw the first lines ganeshew drew was

- dan815

y=x+10
and
x=y+10 right

- Kainui

yeah

- Kainui

notice they're symmetric so we only needed to find one part of that, then multiply by 2

- dan815

ugh its gonna into these inequality cases

- dan815

yeah

- dan815

i kinda see it okay okaokyaok lets dO IT

- dan815

oh i get it

- dan815

we have to consider this stuff, when x+y+k=z
and k>10 or less than 10

- dan815

if k is greater that means x and y are further apart too

- dan815

does that make sense?

- dan815

ok nvm it doesnt threwo it out for now lol

- dan815

weeheeww heeww hooo

- dan815

gotta eat brb

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