33 replies to this topic

[int]

If two babies nap for 1 hour each, and they do so only once a day, what is the probability that on any given day (24 hour period) the babies' naps will overlap. [Assume the time when each nap starts is completely random]. Show all work.

[israeli4ever]

If two babies nap for 1 hour each, and they do so only once a day, what is the probability that on any given day (24 hour period) the babies' naps will overlap. [Assume the time when each nap starts is completely random]. Show all work.

It depends on how tired the mother is that day. The more tired, the less likely.

[Razie]

For the sake of calculation, assume that one minute is the smallest unit of time. Assumption that each baby's hour is chosen randomly.

There are 24*60=1440 minutes in a day.

It does not matter which hour baby 1 sleeps, just take that as a given and calculate the odds that Baby 2 will overlap.*

Naps will overlap if Baby 2's first minute of sleeping is from 59 minutes before baby 1 falls asleep to any of the 60 minutes where baby 1 is already asleep.

59+60 = 119
So there are 119 minutes out of 1440 minutes where baby 2 could fall asleep and overlap with baby 1.
119/1440 =
--------------------
8.26%
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* suppose you bothered mapping all of Baby 1's sleep times. You'd end up adding up the probability of 1440 probabilities, each having the same value, but diminished by the 1/1440 chance that that's when Baby 1 fell asleep. You'd have (1/1440)*8.26% + (1/1440)*8.26% + (1/1440)*8.26% etc etc. It's the same as calculating what are the odds that two babies will be the same sex. You don't have to know what the first one is. Just "What are the odds that the second baby will be the same as the first? = 50%. That's the answer. You'd also get that if you did "Well, odds are 50% that first baby will be a girl and 50% that second will and 50% first will by a boy and 50% that second will, so that answer is 1/2*50% + 1/2*50% = 50% But you don't need to do that.

[Sweet]

If two babies nap for 1 hour each, and they do so only once a day, what is the probability that on any given day (24 hour period) the babies' naps will overlap. [Assume the time when each nap starts is completely random]. Show all work.

I suspect that you are more qualified to answer this question than I am, but I would guess about 1 in 8. But I see this getting complicated.

[TimeRebbe]

never

[int]

For the sake of calculation, assume that one minute is the smallest unit of time. Assumption that each baby's hour is chosen randomly.

There are 24*60=1440 minutes in a day.

It does not matter which hour baby 1 sleeps, just take that as a given and calculate the odds that Baby 2 will overlap.*

Naps will overlap if Baby 2's first minute of sleeping is from 59 minutes before baby 1 falls asleep to any of the 60 minutes where baby 1 is already asleep.

59+60 = 119
So there are 119 minutes out of 1440 minutes where baby 2 could fall asleep and overlap with baby 1.
119/1440 =
--------------------
8.26%
--------------------

* suppose you bothered mapping all of Baby 1's sleep times. You'd end up adding up the probability of 1440 probabilities, each having the same value, but diminished by the 1/1440 chance that that's when Baby 1 fell asleep. You'd have (1/1440)*8.26% + (1/1440)*8.26% + (1/1440)*8.26% etc etc. It's the same as calculating what are the odds that two babies will be the same sex. You don't have to know what the first one is. Just "What are the odds that the second baby will be the same as the first? = 50%. That's the answer. You'd also get that if you did "Well, odds are 50% that first baby will be a girl and 50% that second will and 50% first will by a boy and 50% that second will, so that answer is 1/2*50% + 1/2*50% = 50% But you don't need to do that.

Yeah, except two problems :

1. In your solution, if baby 1 falls asleep at the end of the day, say at hour 23, minute 59, he will only sleep for 1 minute during that day. So then the probability of baby 2 overlapping during that day will be 1/1440. In other words, you can't just simply multiply (59+60)/1440 by 1440, since edge cases are not equal.

2. I never specified that the minute is the smallest unit of time. But generalizing your solution, if we assume the unit of time is such that there are k units in an hour [k=60 in your solution], there will be 24*k units in a day, and baby 2 will overlap with k+(k-1) units (again, ignoring edge cases), so the probability is (2k-1) / 24k. In the limit of k-> inf, it becomes 1/12 = .08333...

[Sweet]

Yeah, except two problems :

1. In your solution, if baby 1 falls asleep at the end of the day, say at hour 23, minute 59, he will only sleep for 1 minute during that day. So then the probability of baby 2 overlapping during that day will be 1/1440. In other words, you can't just simply multiply (59+60)/1440 by 1440, since edge cases are not equal.

2. I never specified that the minute is the smallest unit of time. But generalizing your solution, if we assume the unit of time is such that there are k units in an hour [k=60 in your solution], there will be 24*k units in a day, and baby 2 will overlap with k+(k-1) units (again, ignoring edge cases), so the probability is (2k-1) / 24k. In the limit of k-> inf, it becomes 1/12 = .08333...

But in your new hypo, the baby will not have slept for an hour that day.

[Razie]

Yeah, except two problems :

1. In your solution, if baby 1 falls asleep at the end of the day, say at hour 23, minute 59, he will only sleep for 1 minute during that day. So then the probability of baby 2 overlapping during that day will be 1/1440. In other words, you can't just simply multiply (59+60)/1440 by 1440, since edge cases are not equal.

2. I never specified that the minute is the smallest unit of time. But generalizing your solution, if we assume the unit of time is such that there are k units in an hour [k=60 in your solution], there will be 24*k units in a day, and baby 2 will overlap with k+(k-1) units (again, ignoring edge cases), so the probability is (2k-1) / 24k. In the limit of k-> inf, it becomes 1/12 = .08333...

First, I specified that I was assuming a minute was the smallest time for the sake of calculation. I explicitly acknowledged that I was taking a shortcut that would affect precision.

Even with edge cases and my calculation shortcut, I got to 8.26% which is functionally the right answer. Had I thought you wanted a more precise exact answer I would have gone down that path.

[int]

First, I specified that I was assuming a minute was the smallest time for the sake of calculation. I explicitly acknowledged that I was taking a shortcut that would affect precision.

Yeah, I know you did. It wouldn't fly on an exam though - you can't just make simplifying assumptions that are not present in the problem statement. Or, you can, but then you gotta generalize your solution and still give the right answer.

[Razie]

Yeah, I know you did. It wouldn't fly on an exam though - you can't just make simplifying assumptions that are not present in the problem statement. Or, you can, but then you gotta generalize your solution and still give the right answer.

You didn't suggest this was an exam.

Also, on many exams it WOULD fly. Depends on the class. In many of computer science courses, structuring rough approximations was golden. And simplifying assumptions would get you bonus points for insight - not points detracted. Getting to functional answers by eliminating a rounding error's worth of unnecessary complexity was great.

Also, in real life, probability tends to fly on approximations.

Like, whatever. I thought you wanted an actual functional answer, which I gave you and which was correct. If you wanted to administer a test, you should have let us know, and I wouldn't have played.

[int]

What about if there are 3 babies?

[Xi]

Probability of two overlapping or three?

(null)

[int]

Probability of two overlapping or three?

All three overlapping, for any amount of time.

[Razie]

I will spare you my back-of-the-envelope calculation that took a minute.

[Savannah]

There may be 24 hours in a day, but you have to assume that other than the nap, the babies also sleep (fitfully) through the night.

[Snag]

I can't believe none of you are factoring in Murphy's Law....

[greentiger]

There may be 24 hours in a day, but you have to assume that other than the nap, the babies also sleep (fitfully) through the night.

Yeah i was gonna say that. Sleeping from 2-3 am would not be called a nap. I say you need to cut 6-12 hours out of your 24 hour timeframe.

I can't believe none of you are factoring in Murphy's Law....

It depends on how tired the mother is that day. The more tired, the less likely.

[Savannah]

If two events, A and B are independent then the joint probability is


(1/12)(1/12) = 1/144 = .0069 = .69%

If I'm wrong, please explain why.

[Razie]

If two events, A and B are independent then the joint probability is


(1/12)(1/12) = 1/144 = .0069 = .69%

If I'm wrong, please explain why.

You are wrong because the question is not asking what are the odds that baby 2 and baby 3 both overlap with baby 1. They all have to overlap with each other.

[Savannah]

You are wrong because the question is not asking what are the odds that baby 2 and baby 3 both overlap with baby 1. They all have to overlap with each other.

I'm still on problem #1 in the OP.

But if there are three, why would it not be (1/12)(1/12)(1/12)?