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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: $(10/400)$ + $(3/80) = 1/16$

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    $\begingroup$ What goes wrong when you just write it out in the obvious way? $\endgroup$ – lulu Aug 1 at 18:08
  • $\begingroup$ welcome to MSE. kindly include any attempt. $\endgroup$ – Siong Thye Goh Aug 1 at 18:12
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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 $14$.

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  • $\begingroup$ I've added my approach. Not sure if it's correct. $\endgroup$ – user3119875 Aug 1 at 18:52
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    $\begingroup$ $\frac{1}{2}\times\frac{1}{20}+\frac{3}{20}\times\frac{1}{4}=\frac{1}{16}$ 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. $\endgroup$ – JMoravitz Aug 1 at 18:56
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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.

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