AMC12 2002 B

AMC12 2002 B · Q16

AMC12 2002 B · Q16. It mainly tests Probability (basic), Primes & prime factorization.

Juan rolls a fair regular octahedral die marked with the numbers $1$ through $8$. Then Amal rolls a fair six-sided die. What is the probability that the product of the two rolls is a multiple of 3?

胡安掷一个公平的标有数字$1$到$8$的正八面体骰子。然后阿迈尔掷一个公平的六面骰子。两次掷出的乘积是$3$的倍数的概率是多少？

(A) \frac{1}{12} \frac{1}{12}

(B) \frac{1}{3} \frac{1}{3}

(D) \frac{7}{12} \frac{7}{12}

(E) \frac{2}{3} \frac{2}{3}

Answer

Correct choice: (C)

正确答案：（C）

Solution

On both dice, only the faces with the numbers $3,6$ are divisible by $3$. Let $P(a) = \frac{2}{8} = \frac{1}{4}$ be the probability that Juan rolls a $3$ or a $6$, and $P(b) = \frac{2}{6} = \frac 13$ that Amal does. By the Principle of Inclusion-Exclusion, \[P(a \cup b) = P(a) + P(b) - P(a \cap b) = \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{2} \Rightarrow \mathrm{(C)}\] Alternatively, the probability that Juan rolls a multiple of $3$ is $\frac{1}{4}$, and the probability that Juan does not roll a multiple of $3$ but Amal does is $\left(1 - \frac{1}{4}\right) \cdot \frac{1}{3} = \frac{1}{4}$. Thus the total probability is $\frac 14 + \frac 14 = \frac 12$. The probability that neither Juan nor Amal rolls a multiple of $3$ is $\frac{6}{8} \cdot \frac{4}{6} = \frac{1}{2}$; using complementary counting, the probability that at least one does is $1 - \frac 12 = \frac 12 \Rightarrow \mathrm{(C)}$. 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 $2/8 = 1/4$. The probability that Juan does not roll 3 or 6, but Amal does is $(3/4) (1/3) = 1/4$. Thus, the probability that the product of the rolls is a multiple of 3 is \[\frac{1}{4} + \frac{1}{4} = \boxed{\frac{1}{2}}.\]

在两个骰子上，只有数字$3,6$能被$3$整除。设$P(a)=\frac{2}{8}=\frac{1}{4}$为胡安掷出$3$或$6$的概率，$P(b)=\frac{2}{6}=\frac 13$为阿迈尔掷出$3$或$6$的概率。由容斥原理， \[P(a \cup b)=P(a)+P(b)-P(a \cap b)=\frac{1}{4}+\frac{1}{3}-\frac{1}{4}\cdot\frac{1}{3}=\frac{1}{2} \Rightarrow \mathrm{(C)}\] 或者，胡安掷出$3$的倍数的概率是$\frac{1}{4}$，而胡安没有掷出$3$的倍数但阿迈尔掷出$3$的倍数的概率是$\left(1-\frac{1}{4}\right)\cdot\frac{1}{3}=\frac{1}{4}$。因此总概率为$\frac 14+\frac 14=\frac 12$。两人都没有掷出$3$的倍数的概率是$\frac{6}{8}\cdot\frac{4}{6}=\frac{1}{2}$；用补集计数，至少有一人掷出$3$的倍数的概率是$1-\frac 12=\frac 12 \Rightarrow \mathrm{(C)}$。乘积是$3$的倍数当且仅当两次掷出的结果中至少有一个是$3$或$6$。胡安掷出$3$或$6$的概率是$2/8=1/4$。胡安没有掷出$3$或$6$但阿迈尔掷出的概率是$(3/4)(1/3)=1/4$。因此，乘积是$3$的倍数的概率是 \[\frac{1}{4}+\frac{1}{4}=\boxed{\frac{1}{2}}.\]

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