AMC8 2006

AMC8 2006 · Q17

AMC8 2006 · Q17. It mainly tests Probability (basic), Parity (odd/even).

Jeff rotates spinners P, Q and R and adds the resulting numbers. What is the probability that his sum is an odd number?

杰夫旋转转盘 P、Q 和 R，并将结果数字相加。他的总和是奇数的概率是多少？

(A) $\frac{1}{4}$ $\frac{1}{4}$

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

(D) $\frac{2}{3}$ $\frac{2}{3}$

(E) $\frac{3}{4}$ $\frac{3}{4}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Because the sum of a number from spinner Q and a number from spinner R is always odd, the sum of the numbers on the three spinners will be odd exactly when the number from spinner P is even. Because 2 is the only even number on spinner P, the probability of getting an odd sum is $\frac{1}{3}$.

（B）因为从转盘 Q 取到的数与从转盘 R 取到的数之和总是奇数，所以三个转盘上的数之和为奇数，当且仅当从转盘 P 取到的数是偶数。由于转盘 P 上唯一的偶数是 2，因此得到奇数和的概率是 $\frac{1}{3}$。

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