<s>[INST] A box contains 13 four-sided dice, 36 six-sided dice, and 8 eight-sided dice. A box contains 13 four-sided dice, 36 six-sided dice, and 8 eight-sided dice. One die is chosen at random and then rolled. The probability that the die rolled was a six-sided die, given that the outcome was 5 is / where  and  are relatively prime positive integers. What is +?
On the solutions: ℙ(6∩5)=/ 6⋅57 and ℙ(5)=(36/6)+(8/8)/57.
ℙ(6|5)=4⋅36/ 4⋅36+3⋅8= 6/7.
Could someone explain how the answer is 13? 
I know total dice is 57, but how is the total dice 7/57 and not 44/57. There are 36 six-sided dice and 8 eight-sided dice, and the outcome is 5. So 36 + 8 = 44 or am I forgetting something? Thank you for any help.
 [/INST] There are 57 dice. If one is chosen at random (with uniform probability) then the probability of choosing a six-sided die and rolling a five is:
$\frac{36}{57} \frac{1}{6} = \frac{6}{57}$.
The probability of rolling a 5 in general is:
$\frac{13}{57} \frac{0}{4} + \frac{36}{57} \frac{1}{6} + \frac{8}{57} \frac{1}{8} = \frac{6}{57} + \frac{1}{57} = \frac{7}{57}$
So the conditional probability is:
$\frac{6}{57}$ divided by $\frac{7}{57}$, which is $\frac{6}{7}$.
6 and 7 are coprime, and $6+7=13$
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