Difference between revisions of "2002 AMC 12B Problems/Problem 16"
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\qquad\mathrm{(E)}\ \frac 23</math> | \qquad\mathrm{(E)}\ \frac 23</math> | ||
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
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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=== | + | === Solution 3 (Alcumus) === |
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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> | 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) | ~aopsav (Credit to AoPS Alcumus) | ||
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== 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}} | ||
| − | [[Category:Introductory | + | [[Category:Introductory Combinatorics Problems]] |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 09:26, 5 November 2023
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,
![\[P(a \cup b) = P(a) + P(b) - P(a \cap b) = \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{2} \Rightarrow \mathrm{(C)}\]](http://latex.artofproblemsolving.com/c/c/2/cc29f83d1e2747949dccdb55b4eb258c113092a4.png)
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 ![\[\frac{1}{4} + \frac{1}{4} = \boxed{\frac{1}{2}}.\]](http://latex.artofproblemsolving.com/b/c/2/bc2b91cd826408c90c3ae943e632a2dea937cb89.png)
~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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
