2002 AMC 12B Problems/Problem 16
Problem
Juan rolls a fair regular octahedral die marked with the numbers through . Then Amal rolls a fair six-sided die. What is the probability that the product of the two rolls is a multiple of 3?
Solution
Solution 1
On both dice, only the faces with the numbers are divisible by . Let be the probability that Juan rolls a or a , and that Amal does. By the Principle of Inclusion-Exclusion,
Alternatively, the probability that Juan rolls a multiple of is , and the probability that Juan does not roll a multiple of but Amal does is . Thus the total probability is .
Solution 2
The probability that neither Juan nor Amal rolls a multiple of is ; using complementary counting, the probability that at least one does is .
Solution 3 (Alcumus)
The product will be a multiple of 3 if and only if at least one of the two rolls is a 3 or a 6. The probability that Juan rolls 3 or 6 is . The probability that Juan does not roll 3 or 6, but Amal does is . Thus, the probability that the product of the rolls is a multiple of 3 is ~aopsav (Credit to AoPS Alcumus)
See also
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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All AMC 12 Problems and Solutions |
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