Solution I :
As long as one of the two numbers turn up as 3 or multiple of 3 (in this case, a "6"), the product of the two numbers will be a multiple of 3.
There are 6 * 6 = 36 ways to get 2 numbers. Out of the 36 pairs, you can have
3 - 1
3 - 2
3 - 3
3 - 4
3 - 5
3 - 6, 6 ways.
However, there are only 5 ways left if the other die has a 3 since 3 - 3 only counts as once.
1 - 3
2 - 3
4 - 3
5 - 3
6 - 3
Next we look at "at least one number is "6".
6 - 1
6 - 2
6 - 4 (We already used 6 - 3)
6 - 5
6 - 6 so 5 ways.
The other way around, we have
1 - 6
2 - 6
4 - 6
5 - 6 Total 4 ways, so the answer is
Solution II:
The easiest way to solve this problem is to use complementary counting, which is 1 (100% or total possible way) - none of the the multiples of 3 showing up, so 1 -
Question #2: [2002 AMC-12B #16] Juan rolls a fair regular eight-sided die. Then Amal rolls a fair regular six-sided die. What isthe probability that the product of the two rolls is a multiple of 3?
Solution:
Using complementary counting (see solution II of the previous question), 1 -
