Difference between revisions of "2002 AMC 12B Problems/Problem 16"
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The probability that neither Juan nor Amal rolls a multiple of <math>3</math> is <math>\frac{6}{8} \cdot \frac{4}{6} = \frac{1}{2}</math>; using [[complementary counting]], the probability that at least one does is <math>1 - \frac 12 = \frac 12 \Rightarrow \mathrm{(C)}</math>. | The probability that neither Juan nor Amal rolls a multiple of <math>3</math> is <math>\frac{6}{8} \cdot \frac{4}{6} = \frac{1}{2}</math>; using [[complementary counting]], the probability that at least one does is <math>1 - \frac 12 = \frac 12 \Rightarrow \mathrm{(C)}</math>. | ||
+ | ===Solution 3=== | ||
+ | |||
+ | 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 <math>2/8 = 1/4</math>. The probability that Juan does not roll 3 or 6, but Amal does is <math>(3/4) (1/3) = 1/4</math>. Thus, the probability that the product of the rolls is a multiple of 3 is <cmath>\frac{1}{4} + \frac{1}{4} = \boxed{\frac{1}{2}}.</cmath> | ||
+ | ~aopsav (Credit to AoPS Alcumus) | ||
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=B|num-b=15|num-a=17}} | {{AMC12 box|year=2002|ab=B|num-b=15|num-a=17}} |
Revision as of 20:18, 29 April 2020
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
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 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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