1994 AMC 8 第 22 题

先试着解答 1994 AMC 8 第 22 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1994 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

22.

右图中的两个转盘各转一次,并把得到的两个数相加。两个数之和为偶数的概率是
The two wheels shown at the right are spun and the two resulting numbers are added. The probability that the sum of the two numbers is even is
16\dfrac{1}{6}
14\dfrac{1}{4}
13\dfrac{1}{3}
512\dfrac{5}{12}
49\dfrac{4}{9}
答案:D
知识点:基本概率奇偶性
小提示:
第一个转盘中 33 占一半,11 和 22 各占四分之一;第二个转盘被平均分成三份
On the first wheel 33 covers half and 11 and 22 each cover a quarter; the second wheel is divided into equal thirds
大提示:
和为偶数当且仅当两个数同为偶数或同为奇数
The sum is even when both numbers are even or both are odd
解答:
第一个转盘中,P(1)=14P(1) = \dfrac14,P(2)=14P(2) = \dfrac14,P(3)=12P(3) = \dfrac12。第二个转盘中,44、55、66 各自的概率为 13\dfrac13。
和为偶数需要两个数同奇偶。两数都为奇数时,第一个数是 11 或 33,概率为 (34)\left(\dfrac34\right);第二个数是 55,概率为 (13)\left(\dfrac13\right),所以概率为 34⋅13=14\dfrac34 \cdot \dfrac13 = \dfrac14。两数都为偶数时,第一个数是 22,概率为 (14)\left(\dfrac14\right);第二个数是 44 或 66,概率为 (23)\left(\dfrac23\right),所以概率为 14⋅23=16\dfrac14 \cdot \dfrac23 = \dfrac16。
总概率是 14+16=512\dfrac14 + \dfrac16 = \dfrac{5}{12}。
所以正确答案是 D。
On the first wheel, P(1)=14,P(1) = \dfrac14, P(2)=14,P(2) = \dfrac14, and P(3)=12.P(3) = \dfrac12. On the second wheel, each of 4,4, 5,5, 66 has probability 13.\dfrac13.
The sum is even when both numbers are odd or both are even. Both odd: the first is 11 or 33 (34)\left(\dfrac34\right) and the second is 55 (13),\left(\dfrac13\right), giving 34⋅13=14.\dfrac34 \cdot \dfrac13 = \dfrac14. Both even: the first is 22 (14)\left(\dfrac14\right) and the second is 44 or 66 (23),\left(\dfrac23\right), giving 14⋅23=16.\dfrac14 \cdot \dfrac23 = \dfrac16.
The total probability is 14+16=512.\dfrac14 + \dfrac16 = \dfrac{5}{12}.
Thus, the correct answer is D .
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