AMC12 2015 B

AMC12 2015 B · Q15

AMC12 2015 B · Q15. It mainly tests Basic counting (rules of product/sum), Conditional probability (basic).

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 $\frac{1}{6}$ chance of getting an A in English, and a $\frac{1}{4}$ chance of getting a B. In History, she has a $\frac{1}{4}$ chance of getting an A, and a $\frac{1}{3}$ 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?

在 Rachelle 的学校，A 计 4 分，B 计 3 分，C 计 2 分，D 计 1 分。她四门课的 GPA 是总分除以 4。她确定数学和科学会得 A，英语和历史每门至少 C。她认为英语得 A 的概率是 $\frac{1}{6}$，得 B 的概率是 $\frac{1}{4}$。历史得 A 的概率是 $\frac{1}{4}$，得 B 的概率是 $\frac{1}{3}$，与英语独立。Rachelle 获得 GPA 至少 3.5 的概率是多少？

(A) \frac{11}{72} \frac{11}{72}

(B) \frac{1}{6} \frac{1}{6}

(D) \frac{11}{24} \frac{11}{24}

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

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Rachelle needs a total of at least 14 points to get a 3.5 or higher GPA, so she needs a total of at least 6 points in English and History. The probability of a C in English is $1-\frac{1}{6}-\frac{1}{4}=\frac{7}{12}$, and the probability of a C in History is $1-\frac{1}{4}-\frac{1}{3}=\frac{5}{12}$. The probability that Rachelle earns exactly 6, 7, or 8 total points is computed as follows: 6 points: $\frac{1}{6}\cdot\frac{5}{12}+\frac{1}{4}\cdot\frac{1}{3}+\frac{7}{12}\cdot\frac{1}{4}=\frac{43}{144}$ 7 points: $\frac{1}{6}\cdot\frac{1}{3}+\frac{1}{4}\cdot\frac{1}{4}=\frac{17}{144}$ 8 points: $\frac{1}{6}\cdot\frac{1}{4}=\frac{6}{144}$ The probability that Rachelle will get at least a 3.5 GPA is $\frac{43}{144}+\frac{17}{144}+\frac{6}{144}=\frac{66}{144}=\frac{11}{24}.$

答案（D）：Rachelle 需要总分至少 14 分才能获得 3.5 或更高的 GPA，因此她在英语和历史两科中合计至少需要 6 分。英语得 C 的概率为 $1-\frac{1}{6}-\frac{1}{4}=\frac{7}{12}$，历史得 C 的概率为 $1-\frac{1}{4}-\frac{1}{3}=\frac{5}{12}$。Rachelle 恰好得到 6、7 或 8 个总分点数的概率计算如下： 6 分：$\frac{1}{6}\cdot\frac{5}{12}+\frac{1}{4}\cdot\frac{1}{3}+\frac{7}{12}\cdot\frac{1}{4}=\frac{43}{144}$ 7 分：$\frac{1}{6}\cdot\frac{1}{3}+\frac{1}{4}\cdot\frac{1}{4}=\frac{17}{144}$ 8 分：$\frac{1}{6}\cdot\frac{1}{4}=\frac{6}{144}$ Rachelle 获得至少 3.5 GPA 的概率为 $\frac{43}{144}+\frac{17}{144}+\frac{6}{144}=\frac{66}{144}=\frac{11}{24}.$

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