AMC10 2008 B

AMC10 2008 B · Q16

AMC10 2008 B · Q16. It mainly tests Probability (basic).

Two fair coins are to be tossed once. For each head that results, one fair die is to be rolled. What is the probability that the sum of the die rolls is odd? (Note that if no die is rolled, the sum is 0.)

抛掷两枚公平硬币一次。每次出现正面，就抛掷一枚公平骰子。骰子点数之和为奇数的概率是多少？（注意：如果没有骰子被抛掷，和为 0。）

(A) \frac{3}{8} \frac{3}{8}

(B) \frac{1}{2} \frac{1}{2}

(D) \frac{5}{8} \frac{5}{8}

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

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): If one die is rolled, 3 of the 6 possible numbers are odd. If two dice are rolled, 18 of the 36 possible outcomes have odd sums. In each of these cases, the probability of an odd sum is $\frac{1}{2}$. If no die is rolled, the sum is 0, which is not odd. The probability that no die is rolled is equal to the probability that both coin tosses are tails, which is $\left(\frac{1}{2}\right)^2=\frac{1}{4}$. Thus the requested probability is \[ \left(1-\frac{1}{4}\right)\cdot\frac{1}{2}=\frac{3}{8}. \]

答案（A）：如果掷一个骰子，6 个可能结果中有 3 个是奇数。如果掷两个骰子，36 个可能结果中有 18 个点数和为奇数。在这些情况下，点数和为奇数的概率都是 $\frac{1}{2}$。如果不掷骰子，则和为 0，而 0 不是奇数。不掷任何骰子的概率等于两次抛硬币都为反面的概率，即 $\left(\frac{1}{2}\right)^2=\frac{1}{4}$。因此所求概率为 \[ \left(1-\frac{1}{4}\right)\cdot\frac{1}{2}=\frac{3}{8}. \]

Topics

CP.006 Probability (basic)