How do you do 185 times 12 in standard algorithm?
1 answer:
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Step-by-step explanation:
Sample space, S = {(1,1),(1,3),(1,4),(1,5),(1,6),(1,8),(2,1),(2,3),(2,4),(2,5),(2,6),(2,8),(2,1),(2,3),(2,4),(2,5),(2,6),(2,8),(3,1),(3,3),(3,4),(3,5),(3,6),(3,8),(3,1),(3,3),(3,4),(3,5),(3,6),(3,8),(4,1),(4,3),(4,4),(4,5),(4,6),(4,8)}
These dice will give a sum of 2 with N={(1,1)}, which only has 1 combination
Thus, the probability of rolling a sum of 2 with these dice is