Question

Find the probability of rolling a twelve first and then a three when a pair of dice is rolled twice?

Answer 1

Total number of possible outcomes = 36

Total number of outcomes with a sum of 12 = 1
{6,6)

Total number of outcomes with a sum of 3 = 2
{1,2}{2,1}

P(12 first, then 3) = (1/36)(2/36) = 1/648

Answer: The probability is 1/648

Answer 2

$|\Omega|=6^2\cdot6^2=1296\\ |A|=1\cdot2=2\\\\ P(A)=\dfrac{2}{1296}=\dfrac{1}{648}\approx0.2\%$