The minute hand of a watch jumps 0, 1 or 2 minutes every min. Find (approx.) the probability that after 2.5 hours, the watch is late by 5 minutes.

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.

$E[X]=\frac{1+2+0}{3}=1$

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

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

$$E[X^2]=\frac{4+1+0}{3}=\frac{5}{3}$$

$$Var(X)=\frac{5}{3}-\frac{3}{3}=\frac{2}{3}$$

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

$$Pr\left(\sum_{i=1}^{150}X_i\ge155\right)$$

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?

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

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 $5$.
$$\frac{\sum\limits_{n=5}^{150}\sum\limits_{k=0}^{150-n}\binom{150}{k}\cdot\binom{150-k}{k+n}}{3^{150}}\approx32.6477\%$$