When two unbiased dice are rolled one by one, what is the probability that either the first one is $2$ or the sum of the two is less than $5$? 
When two unbiased dice are rolled one by one, what is the probability that either the first one is $2$ or the sum of the two is less than $5$?
a) $\dfrac 16$
b) $\dfrac 29$
c) $\dfrac 5{18}$
d) $\dfrac 13$
(Not homework, I'm doing some mock exams I found online)

Now I was pretty sure I got this. Take probability of rolling $2$ on the first one $\left(\frac 16\right)$ and the probability that the sum is $< 5 \left(\frac 6{36}\right)$, add them together and bam, $\frac 13$.
Wrong, seems the answer is c) $\frac5{18}$.
My problem is I've been trying to brush on my probability skills by googling and can see problems I made, such as the wording meaning that rolling a $2$ AND having a total of $5$ NOT being included in the result, but no matter what angle I attack from, I never end up with $\frac5{18}$.
So which little thing did I miss? Thanks very much for helping.

You need to calculate the probability that either the first die is $2$, OR the sum is less than $5$. In math, "OR" always includes the possibility that both things occur (i.e. "either A or B" includes when A and B both happen). 
However, you added the probability that the first die was $2$ ($6/36$) and the probability that the sum was less than $5$ ($6/36$)-- that means that in the case that both occur, you counted it twice, since it will be included in both first term and the second term. This is called double-counting. To fix double-counting, you need to subtract off what you counted twice.
In this case, what you counted twice is when the first die is $2$ AND the sum is less than $5$ -- this happens if the two dice are $(2,1)$ or $(2,2)$, so the probability is $2/36$. Your final answer will be
$$
\frac{6}{36} + \frac{6}{36} - \frac{2}{36} = \frac{10}{36} = \frac{5}{18}.
$$
