2015 AMC 12B Problem 15

Question

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

A

3.5

GPA needs

14

total points, so English and History must give at least

6

points

Find each class’s chance of a C, then add the probabilities of totals

6,

7,

and

8

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.

Problem 15 in other years