Question

在 Rachelle 的学校，A 记 $4$ 分，B 记 $3$ 分，C 记 $2$ 分，D 记 $1$ 分。她所修四门课的 GPA 是总分除以 $4$ 。她确定自己在数学和科学两科都会得 A，并且英语和历史每科至少得 C。她认为自己英语得 A 的概率是 $\dfrac16$ ，得 B 的概率是 $\dfrac14$ 。历史得 A 的概率是 $\dfrac14$ ，得 B 的概率是 $\dfrac13$ ，且与英语成绩相互独立。Rachelle 的 GPA 至少为 $3.5$ 的概率是多少？

At Rachelle’s school an A counts $4$ points, a B $3$ points, a C $2$ points, and a D $1$ point. Her GPA on the four classes she is taking is computed as the total sum of points divided by $4.$ She is certain that she will get As in both Mathematics and Science, and at least a C in each of English and History. She thinks she has a $\dfrac16$ chance of getting an A in English, and a $\dfrac14$ chance of getting a B. In History, she has a $\dfrac14$ chance of getting an A, and a $\dfrac13$ chance of getting a B, independently of what she gets in English. What is the probability that Rachelle will get a GPA of at least $3.5?$

$\dfrac{11}{72}$

$\dfrac16$

$\dfrac{3}{16}$

$\dfrac{11}{24}$

$\dfrac12$

答案：D

知识点：独立事件分类讨论

难度评级：1820

小提示：

$3.5$ 的 GPA 需要总分 $14$ ，所以英语和历史至少要给 $6$ 分

A $3.5$ GPA needs $14$ total points, so English and History must give at least $6$ points

大提示：

求出每科得 C 的概率，再加上总分为 $6$ ， $7$ ，和 $8$ 的概率

Find each class’s chance of a C, then add the probabilities of totals $6,$ $7,$ and $8$

解答：

数学和科学给 $8$ 分，所以 Rachelle 还需要英语和历史至少给 $6$ 分。英语得 C 的概率是 $1 - \dfrac16 - \dfrac14 = \dfrac{7}{12}$ ，历史得 C 的概率是 $1 - \dfrac14 - \dfrac13 = \dfrac{5}{12}$ 。

统一到分母 $144$ ： $8$ 分的概率为 $\dfrac16\cdot\dfrac14 = \dfrac{6}{144}$ ； $7$ 分的概率为 $\dfrac16\cdot\dfrac13 + \dfrac14\cdot\dfrac14 = \dfrac{17}{144}$ ； $6$ 分的概率为 $\dfrac16\cdot\dfrac{5}{12} + \dfrac14\cdot\dfrac13$ $+ \dfrac{7}{12}\cdot\dfrac14 = \dfrac{43}{144}$ 。

总概率为 $\dfrac{6 + 17 + 43}{144} = \dfrac{66}{144} = \dfrac{11}{24}$ 。

因此，正确选项是 D。

Math and Science give $8$ points, so Rachelle needs at least $6$ more from English and History. The chance of a C is $1 - \dfrac16 - \dfrac14 = \dfrac{7}{12}$ in English and $1 - \dfrac14 - \dfrac13 = \dfrac{5}{12}$ in History.

Working over a denominator of $144:$ $8$ points has probability $\dfrac16\cdot\dfrac14 = \dfrac{6}{144};$ $7$ points has $\dfrac16\cdot\dfrac13 + \dfrac14\cdot\dfrac14 = \dfrac{17}{144};$ and $6$ points has $\dfrac16\cdot\dfrac{5}{12} + \dfrac14\cdot\dfrac13$ $+ \dfrac{7}{12}\cdot\dfrac14 = \dfrac{43}{144}.$

The total is $\dfrac{6 + 17 + 43}{144} = \dfrac{66}{144} = \dfrac{11}{24}.$

Thus, the correct answer is D.

2015 AMC 12B 第 15 题

答案：D

难度评级：1820

小提示：

大提示：

解答：

其他年份的第 15 题