Joe Should Go To Bed

If Joe goes to sleep at a random time between 10 pm and 11 pm, and sleep for a random length of time between 8 to 10 hours, What is the probability that he wakes up before 8:30 am?

@joe-shmo said
If Joe goes to sleep at a random time between 10 pm and 11 pm, and sleep for a random length of time between 8 to 10 hours, What is the probability that he wakes up before 8:30 am?

It will depend on whether it is the night that the clocks are adjusted for daylight saving!
All I can say for now is to guarantee waking before 08:30 am(prob =1) and not knowing how long you will sleep for you must go to sleep by 10:29pm
Sleep well.

@venda said
It will depend on whether it is the night that the clocks are adjusted for daylight saving!

Its not! No tricks like that.

@joe-shmo said
Its not! No tricks like that.

O.K mate.
I've edited my post above, but I'm sure that's not the answer either!!

@venda said
It will depend on whether it is the night that the clocks are adjusted for daylight saving!
All I can say for now is to guarantee waking before 08:30 am(prob =1) and not knowing how long you will sleep for you must go to sleep by 10:29pm
Sleep well.

Well, that's actually a step in the right direction. Except he doesn't have to go to bed at 10:29, he could go to bed at 10:29:59 etc... The time he has to go to bed has to be: T ≤ 10:30 to wake up at most "exactly" at 8:30 ( the probability of waking "exactly" at 8:30 is actually 0 ).

Hint:
Reveal Hidden Content

So my attempt:

Reveal Hidden Content

A lot of edits to get the hidden thingie right...🙁

@Ponderable

For some reason we cant hide multiline text with "returns".

Anyhow, your answer is in the ball park. I'm having trouble visualizing how you have arrived to determine where the discrepancy may lie.

It sounds like you are doing some type of numerical approximation ( Riemann Sum ) where the smallest unit of time is 1 minute. If that is the case, I suppose that is ok ( I was hoping for the analytical approach ). But if that is the case there is typically going to be some upper and lower bound associated with the numerical solution.

Have you given the lower or upper bound, what is the the other boundary and justification for the boundary magnitude?

It seems like you have taken the 2 hr long range, divided it up into minutes ( 120 minutes ) and concluded that the probability of waking up in any given minute is 1/120. Is that an accurate assessment, or am I not getting what you are doing?

@joe-shmo said
If Joe goes to sleep at a random time between 10 pm and 11 pm, and sleep for a random length of time between 8 to 10 hours, What is the probability that he wakes up before 8:30 am?

This can be seen as the sum of two independent uniformly distributed random variables.

V1 ~ U(0,1)
V2 ~U(8,10)

sum distribution = 0 + (1-0)u1 + 8 + (10-8)u2.

You have to find where this distribution is less than 10.5:

8 + u1 + 2(u2) < 10.5

Which simplifies to (1/2)(u1) + u2 < 5/4

The area under the curve y = (-1/2)x + 5/4 within a square on the Cartesian plane bounded by the points (0,0); (0,1); (1,1) and (1,0) = 0.9375.

@joe-shmo said
@Ponderable

For some reason we cant hide multiline text with "returns".

Anyhow, your answer is in the ball park. I'm having trouble visualizing how you have arrived to determine where the discrepancy may lie.

It sounds like you are doing some type of numerical approximation ( Riemann Sum ) where the smallest unit of time is 1 minute. If that is the case, I suppose ...[text shortened]... e lower or upper bound, what is the the other boundary and justification for the boundary magnitude?

His slight error comes from the fact that he is effectively treating this as a discrete probability distribution when it is continuous, and thus requires a density function.

Riemann sums of course lead to varying degrees of error depending on how wide the columns are.

@ashiitaka said
This can be seen as the sum of two independent uniformly distributed random variables.

V1 ~ U(0,1)
V2 ~U(8,10)

sum distribution = 0 + (1-0)u1 + 8 + (10-8)u2.

You have to find where this distribution is less than 10.5:

8 + u1 + 2(u2) < 10.5

Which simplifies to (1/2)(u1) + u2 < 5/4

The area under the curve y = (-1/2)x + 5/4 within a square on the Cartesian plane bounded by the points (0,0); (0,1); (1,1) and (1,0) = 0.9375.

Correct! ( I think - its uber apparent your more adept in probability than myself )

Since this is a uniform distribution:
I solved by making a plot of time asleep vs time awake. All possible outcomes are represented by the area a parallelogram with vertices ( 10,6 ), ( 10,8 ), (11,7), (11,9 )

This parallelogram has a base of 2 (hr) and a height of 1 (hr).

Waking after 8:30 The area of the parallelogram to the right of 8:30 ( a right triangle ) divided by the area of the entire parallelogram.

P ( Waking > 8:30 ) = (1/2)^3 /2 = (1/2)^4

Finally:

P ( Waking < 8:30 ) = 1 - (1/2)^4 = 15/16

@joe-shmo said
Correct! ( I think - its uber apparent your more adept in probability than myself )

Since this is a uniform distribution:
I solved by making a plot of time asleep vs time awake. All possible outcomes are represented by the area a parallelogram with vertices ( 10,6 ), ( 10,8 ), (11,7), (11,9 )

This parallelogram has a base of 2 (hr) and a height of 1 (hr).

Waking ...[text shortened]...
P ( Waking > 8:30 ) = (1/2)^3 /2 = (1/2)^4

Finally:

P ( Waking < 8:30 ) = 1 - (1/2)^4 = 15/16

Shows there is usually more than 1 way to approach a problem.
I vow to spend time analysing the various problems to see how much I can understand!
Maybe if there's another lock down?

@venda said
Shows there is usually more than 1 way to approach a problem.
I vow to spend time analysing the various problems to see how much I can understand!
Maybe if there's another lock down?

Maybe if there's another lock down?

...Even if it gives extra time to solve puzzles, surely one would hope not!