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?
 A: You did not take into account the fact that a 2 or a 5 could be obtained in either the first roll or the second roll.  You must add these probabilities.  Also, you must take into account the possibility that both events occur.
The probability of obtaining a 2 on the first roll is $1/6$.  The probability of obtaining a 2 on the second roll is also $1/6$.  Similarly, the probability of obtaining a 5 on the first roll is $1/6$, and the probability of obtaining a 5 on the second roll is also $1/6$.  Adding those four probabilities yields 
$$4 \cdot \frac{1}{6} = \frac{2}{3}$$
However, we have counted those outcomes in which we obtain a 2 on both rolls, a 5 on both rolls, a 2 on the first roll and a 5 on the second roll, and a 5 on the first roll and a 2 on the second roll twice.  Each of those four events has probability $1/36$.  Hence, the probability of obtaining a 2 or a five in two throws of a die is 
$$4 \cdot \frac{1}{6} - 4 \cdot \frac{1}{36}  = \frac{2}{3} - \frac{1}{9} = \frac{5}{9}$$
as you found by listing the possibilities.  
A: 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.
A: I presume that if you know the probability rule for addition, you also know that for multiplication, often called the fundamental counting principle
You need either a 2, or a 5 or both.
In such situations, use of the complement can greatly simplifies computations.
The outcome we don't want, P(neither $2$ nor $5$),
thus $P(2\;or\;5) = 1 - \dfrac46\cdot\dfrac46 = \dfrac{20}{36}= \dfrac59$
