# 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 $$<= 10$$ {$$(10,4),(9,5),(8,6),(7,7),(6,8),(5,9),(4,10),(3,11),(2,12),(1,13)$$} and $$3$$ ways we can get sum of $$14$$ if the first dice has value $$>= 11$$ {$$(11,3),(12,2),(13,1)$$} so the probability is: $$(10/400)$$ + $$(3/80) = 1/16$$

• What goes wrong when you just write it out in the obvious way? – lulu Aug 1 at 18:08
• welcome to MSE. kindly include any attempt. – Siong Thye Goh Aug 1 at 18:12

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 $$14+$$.

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 $$14$$.

• I've added my approach. Not sure if it's correct. – user3119875 Aug 1 at 18:52
• $\frac{1}{2}\times\frac{1}{20}+\frac{3}{20}\times\frac{1}{4}=\frac{1}{16}$ is correct. There is absolutely no need to write out each of the possibilities however and this causes errors when trying to count things by hand and wastes effort and time. – JMoravitz Aug 1 at 18:56

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

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