Kainui
  • 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?
Probability
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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rational
  • rational
|dw:1432999492312:dw|
Kainui
  • Kainui
Hahaha yeah you got it.
rational
  • rational
P(|E1-E2| > 10 minutes) = (Area of shaded region) / (Total area) = 5^2/6^2 = 25/36 ?

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Kainui
  • Kainui
Yep, exactly. =)
rational
  • 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
  • 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
  • rational
very interesting, 3 events requires a triple integral is it
rational
  • rational
second part of the question is easy yeah we always get squares in top and bottom because of symmetry : |E1-E2| > 20
Kainui
  • Kainui
for the 2D case in general the answer is: \[\frac{(60-t)^2}{60^2}\] which is quite nice!
ParthKohli
  • ParthKohli
Wow, that's beautiful.
ParthKohli
  • ParthKohli
Looks like my book has this... is this what is called "infinitistic probability"?
Kainui
  • Kainui
Yup, I guess? I guess continuous probability distribution maybe? It's important for quantum mechanics!
ikram002p
  • ikram002p
:)
dan815
  • dan815
|dw:1433020038430:dw|
dan815
  • dan815
oh greater so 5/6
dan815
  • dan815
|dw:1433020148264:dw|
dan815
  • dan815
is that it?
dan815
  • 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
  • dan815
i solved it by looking at a number line
dan815
  • dan815
|dw:1433020285038:dw|
dan815
  • 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
  • dan815
Hence 5/6 for the complete thing
dan815
  • dan815
here is also something interesting... prolly not so much but kinda cool to visualze lol
dan815
  • dan815
cause its not good to think ofa number line but a circular number like for this or like a clock
dan815
  • dan815
|dw:1433020557004:dw|
dan815
  • dan815
|dw:1433020604712:dw|
Kainui
  • Kainui
the problem is your answer seems to be wrong, the answer is 25/36 not 5/6 for the first question
dan815
  • dan815
ya i got that i change it xD
dan815
  • dan815
for a sec i thought u were asking less than 10
dan815
  • dan815
oh dang really
dan815
  • dan815
why is it 5^/6^2?
dan815
  • dan815
lemmee think
Kainui
  • 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
  • dan815
oh :O i get it
dan815
  • dan815
theres a problem here with assuming the circular thing too, because once event 1 occurs later
dan815
  • dan815
the prob is greater for 10 mins
dan815
  • dan815
:D
dan815
  • dan815
or wait no, the even can happen back in time too
Kainui
  • Kainui
This is pretty similar to the Buffon needle problem which is a cool way to approximate \(\pi\).
dan815
  • dan815
i dont think i understand the graph
Kainui
  • 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
  • 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
  • dan815
ah i gotcha now xD
dan815
  • dan815
the graph labelling didnt quite make sense to me, now its clear thanks
Kainui
  • Kainui
cool so what if we have 3 events separated by 10 minutes, what does that cube look like?
dan815
  • dan815
|dw:1433021555908:dw|
Kainui
  • 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
  • dan815
ok lemme see more algrebrically first a+b+c<60*3 and diff between a,b,c is greater than 10
dan815
  • dan815
(5/6)^3?
dan815
  • dan815
^ thats a guess
dan815
  • 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
  • Kainui
|dw:1433021690423:dw|
dan815
  • 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
  • Kainui
So one example of an event can be like (10, 20, 30) since they are all separated by 10 min.
dan815
  • dan815
|dw:1433021882772:dw|
Kainui
  • Kainui
yeah what I drew is wrong I think idk haha
Kainui
  • Kainui
Yeah I just didn't wanna draw it with those lines in there cause it looked tacky
dan815
  • dan815
|dw:1433021915541:dw|
dan815
  • dan815
lol
dan815
  • dan815
oh u can think of the 2by 2 case as 1 big event again
dan815
  • dan815
|dw:1433022504943:dw|
dan815
  • dan815
so u solve for cases, when event 1 and 2 are close, and when they are not
dan815
  • dan815
eh nvm i feel like there is something easier to compute off the differences
dan815
  • dan815
did u already solve it?
Kainui
  • Kainui
Nahhhh
dan815
  • dan815
ok :) im on it then sir
Kainui
  • 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
  • dan815
that what i was lookign at
dan815
  • dan815
like can we write out system of equationa for lines
dan815
  • dan815
and break it down into 4 or 6 or 9 cases or something
Kainui
  • Kainui
I think this is enough to make up an integral, those are our bounds
dan815
  • dan815
hey btw the first lines ganeshew drew was
dan815
  • dan815
y=x+10 and x=y+10 right
Kainui
  • Kainui
yeah
Kainui
  • Kainui
notice they're symmetric so we only needed to find one part of that, then multiply by 2
dan815
  • dan815
ugh its gonna into these inequality cases
dan815
  • dan815
yeah
dan815
  • dan815
i kinda see it okay okaokyaok lets dO IT
dan815
  • dan815
oh i get it
dan815
  • dan815
we have to consider this stuff, when x+y+k=z and k>10 or less than 10
dan815
  • dan815
if k is greater that means x and y are further apart too
dan815
  • dan815
does that make sense?
dan815
  • dan815
ok nvm it doesnt threwo it out for now lol
dan815
  • dan815
weeheeww heeww hooo
dan815
  • dan815
gotta eat brb

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