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How can I find the probability of waking up at a precise minute? Say I fall asleep at $10$ PM and wake up at $6:01$ AM. There's a total of $481$ minutes we are dealing with so the odds of waking up at an exact minute would be $1:480$ right? And the probability would be $0.2$% ($1/481$), correct? But what about external factors. Like what if we didn't know the exact amount of time I would be asleep? Or what if there was a noise that woke me up in the middle of the night. How would these things be added to the equation if you were trying to find the probability of someone waking up at an exact minute?

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    I don't think this is a math problem. Surely the probability at $10$:$01$ is not the same as it is at $5$:$59$. Nor have we got any information about whatever noise might occur (is there an alarm clock in the picture? That would certainly increase the probability for whenever it is set). – 2017-01-02
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    Your hypothesis is completely wrong. Using this reasoning, the probability to wake up at $10:01$ PM is... $1$. Not talking about the probability to wake up at $10:00$... – 2017-01-02

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For the average British population, the duration of the sleep is well modeled by a Gaussian with $\mu=7.04$h and $\sigma=1.55$h [Groeger et al. 2004]. To estimate the probability to wake-up after $m$ minutes,

$$W_m:=\mathbb P(m\le t

where $\text{erf}$ denotes the error function.

This is well approximated by

$$\frac{e^{-(m-\mu+1/2)^2/2\sigma^2}}{\sqrt{2\pi}\sigma}.$$

If we model the occurence of a nightly wakening noise with an exponential law,

$$N_m:=\mathbb P(t

(say $\lambda=(\ln2/3.5)h^{-1}$ so that it occurs in the middle of the night with probability $1/2$) the probability to wake-up is the product of the probability to spontaneously wake-up in the minute $m$ by the probability of not having been woken-up earlier, i.e.

$$W'_m:=W_m(1-N_m)=\frac12\left(\text{erf}\left(\frac{m+1-\mu}{\sqrt2\sigma}\right)-\text{erf}\left(\frac{m-\mu}{\sqrt2\sigma}\right)\right)e^{-\lambda m}.$$

If I am right, under these assumptions,

$$W_{481'}=0.003505,\\ N_{481'}=0.760807,\\ W'_{481'}=0.000717.$$

Note that $W_m$ nearly follows a Gaussian, which is damped by an exponential, giving for $W'_m$ another Gaussian which if a shifted version of the former. So the effect of the noise is to shorten the average sleep duration by $\lambda\sigma^2\approx 28.5$ minutes.

Finally, note that this development is valid for the "average British". The constants and even the distribution can be different for a particular individual, for instance using an alarm clock (resulting in a truncated Gaussian).

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    "For the average British population, the duration of the sleep is well modeled by a Gaussian..." Interesting! Where can I read more about this study? – 2017-01-02
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    @Rahul: Google "statistical distribution of sleep duration". – 2017-01-02
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    Ah, neat, I didn't know there was this much literature out there! I would like to edit the reference into your answer but I don't know which of the many search hits was the one you used. – 2017-01-02
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Suppose that "waking up" is a Poisson process (just as the process of the arrival of a customer at a shop) with parameter T then the probability of waking up after a sleeping time t is

$$w(t)=1-\exp \left(-\frac{t}{T}\right)$$

The PDF of the waking-up time t, i.e. the probability density of waking up between $t$ and $t + dt$, is

$$f(t)=\frac{\exp \left(-\frac{t}{T}\right)}{T}$$

The average waking-up time is $T$.

Extension

You could also ask for the probabilty of waking up $n$ times in the interval between $0$ and $t$. This leads to a Poisson distribution as the reader can verify.