probability of rolling a sum of $14$ when rolling a $20$-sided die twice or a $20$-sided die with a $4$-sided die depending on outcome of first roll A $20$-sided die is rolled. If the result is $10$ or less, the $20$-sided die is rolled again. If the result is $11$ or more, a $4$-sided die is rolled. In either case, the results are summed after the second die roll. What is the probability that the sum will be $14$?

My Approach:
there are 10 ways we can get a sum of 14 if the first dice is  $&lt;= 10$ {$(10,4),(9,5),(8,6),(7,7),(6,8),(5,9),(4,10),(3,11),(2,12),(1,13)$} and $3$ ways we can get sum of $14$ if the first dice has value  $&gt;= 11$ {$(11,3),(12,2),(13,1)$} so the probability is: $(10/400)$ + $(3/80) = 1/16$
Hints:

$Pr(A\cap B) = Pr(A)Pr(B\mid A)$

$Pr(B) = Pr(A_1\cap B)+Pr(A_2\cap B)+\dots+Pr(A_n\cap B)$ where $A_1,\dots,A_n$ form a partition of the sample space



Let $B$ be the event the sum is $14$.

Let $A_1$ be the event the first die rolls a number $1-10$.  Let $A_2$ be the event the first die rolls a number $11-13$.  Let $A_3$ be the event the first die rolls a number $14+$.

Additional Hint:


   Regardless what was rolled on the first die, so long as it was less than $14$, there will be exactly one outcome on the second roll that would let the total sum to $14$.
The probability that you obtain $14$ using the $20$-sided die on the second role is 

$$\frac{10}{20} \cdot \frac{1}{20}$$

The probability you obtain $14$ using the $4$-sided die on the second role is

$$\frac{3}{20} \cdot \frac{1}{4}$$

Add these.