The mind hand of an old watch jumps 0, 1 or 2 minutes every minute with a probability of 1/3 for each value, independent of one another.. Find (approx.) the probability that after 2.5 hours, the watch is late by 5 minutes or more.

So we need to use the central limit theorem:

I thought this was a uniform distribution, but since it takes on discrete value (0,1,3) it isn't. Therefore the mean is found by adding the values and dividing by n values.


But I'm having two issues. 1.) Finding the correct Var(X)

$Var[X]=E[X^2]-E^2[X]$= $E[X^2]-1$



And since we want the outcome after after 150 trials(each jump) we want


Can anyone double check my work for Var(X) and explain why the Probability I've listed above isn't correct?

EDIT: I've made a mistake in understanding the correct translation of the question. It wanted 5 or more minutes late i.e. 145 instead of 150. That clears that part up, but is the calculation of the variance correct?

  • $\begingroup$ Should add "or more" in the title. $\endgroup$ Aug 4, 2016 at 9:49
  • $\begingroup$ @barakmanos had no room to properly add all the details of the question, so I added them in the body of the post. $\endgroup$
    – RonaldB
    Aug 4, 2016 at 9:50
  • $\begingroup$ So change the title entirely to something more generic (seems better than putting technical details, and only some of them). $\endgroup$ Aug 4, 2016 at 9:50

1 Answer 1


Let's rephrase the question:

Given a sequence of $150$ uniformly distributed elements $\in\{0,1,2\}$ , find the probability that the amount of $2$s is larger than the amount of $0$s by at least $5$.

We can solve it by splitting it into disjoint events, and then add up their probabilities:


  • $\begingroup$ A little unsure of this (alternative) solution, let's wait for some feedback... $\endgroup$ Aug 4, 2016 at 10:03
  • $\begingroup$ The solution is off by ~0.02% from a solution using the central limit theory. $\endgroup$
    – RonaldB
    Aug 4, 2016 at 10:15
  • $\begingroup$ @true blue anil: Can you please review? $\endgroup$ Aug 6, 2016 at 5:53
  • $\begingroup$ @N. F. Taussig: Can you please review? $\endgroup$ Aug 6, 2016 at 5:53

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