probability of getting 2 or 5 in two throws of a die So I know the probability rule of addition. 
Getting 2 or 5 in two throws should be 
$P(2)+ P(5)$.  $P(2) = 1/6, P(5) = 1/6$ so the combined so it should be $1/3$. 
I tried to visualize but not able to do so correctly.
11,12,13,14,15,16, 21,22,23,24,25,26,31,32, ....6,6
total of $36$ possibilities.
12,15,21,22,23,24,25,26,31,35,42,45,51,52,53,54,55,56,61,65
out of which $20$ possibilities, so the probability should be $20/36$ which is not $1/3$.
Where am I going wrong?

Your both methods have some mistake.

Method 1 -

Getting 2 on  first number and any other number on second except 5. But 2 can be on second die also. So multiply cases with 2.
$2 \left(\frac16 \times \frac46\right)$
= $\left(\frac13 \times \frac23\right)$
= $\left(\frac29\right)$
Similarly for getting 5 on  first number and any other number on second except 2. But 5 can be on second die also. So multiply cases with 2.
$2 \left(\frac16 \times \frac46\right)$
= $\left(\frac13 \times \frac23\right)$
= $\left(\frac29\right)$
Case with 2 on both dice or 5 on both dice.
$\left(\frac16 \times \frac16 + \frac16 \times \frac16\right)$
= $\left(\frac1{36} + \frac1{36}\right)$
= $\left(\frac2{36}\right)$
Case with 2 on first die and 5 on second or vice versa.
$\frac16 \times \frac16 + \frac16 \times \frac16$
= $\left(\frac1{36} + \frac1{36}\right)$
= $\left(\frac2{36}\right)$
Combining these,
$\left(\frac29 + \frac29 + \frac2{36} + \frac2{36}\right)$
= $\left(\frac{20}{36}\right)$

Method 2 -

21, 22, 23, 24, 25, 26, 12, 32, 42, 52, 62, 15, 35, 45, 55, 65, 51, 53, 54, 56 = 20 cases.
