What is the probability of getting the sum of 5 or at least one 4 when you roll a dice? I just want to know if my method is right:
P(Sum of 5 or At least one 4) = 2+3, 3+2, 4+1, 1+4 [+] (4+1,4+2,4+3,4+4,4+5,4+6)*2
So that will be 4+12/36
Ans: 16/36
am i right here?
 A: Note that each of $(4, 1)$ and $(1,4)$ appear twice in each of the cases that you sum (those combinations are counted in the combinations that total $5$, and they are counted twice in the combinations in which at least one $4$ appears, so you're currently double counting each of those. You're also counting $(4, 4)$ twice, when doubling (to account for the permutation of) the combinations in which at least one $4$ appears. So you'd have to subtract a total of $3$ from your current total in the numerator:
$$\frac{4 + 12 - 3}{36} = \frac{[4 + (2\cdot 5 + 1)] - 2}{36} = \frac{13}{36}$$
A: Making a list and carefully counting is a good idea. Even though it is not really necessary in this case, we will look at things in a more abstract way. Let $A$ be the event "sum is $5$" and let $B$ be the event "at least one $4$." Depending on whether you are in a counting mood or in a probability mood, we have 
$$|A\cup B| =|A|+|B|-|A\cap B|,\tag{1}$$ 
or 
$$\Pr(A\cup B) =\Pr(A)+\Pr(B)-\Pr(A\cap B).\tag{2}$$
In (1), we are using $|X|$ to denote the number of elemements in the set $X$. Your course may use a different notation.
The nice thing about the above formulas is that they can help organize the calculations.
We want to find $|A\cup B|$, and then divide by $36$ to find the probability. Or else we can use (2) directly. We will use (1), since it is more closely connnected with what you did.
First we find $|A|$, the number of ways we can have a sum of $5$. You made the count, it is $4$.
Next we find $|B|$, the number of ways to have at least one $4$. Note that there are $5^2$ ways to have no $4$, so there are $36-25=11$ ways to have at least one $4$. Or else we could list and count directly.
Next we find $|A\cap B|$. We want a sum of $5$ and at least one $4$. There are $2$ ways to do this.
Now from (1) we have $|A\cup B|=4+11-2$.  
