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2015 AMC 12B Problem 15

Problem 15 of 25IntermediateProbability & StatisticsProblem-Solving Techniques

At Rachelle’s school an A counts 44 points, a B 33 points, a C 22 points, and a D 11 point. Her GPA on the four classes she is taking is computed as the total sum of points divided by 4.4. 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 16\dfrac16 chance of getting an A in English, and a 14\dfrac14 chance of getting a B. In History, she has a 14\dfrac14 chance of getting an A, and a 13\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?3.5?

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Solution

Math and Science give 88 points, so Rachelle needs at least 66 more from English and History. The chance of a C is 1−16−14=7121 - \dfrac16 - \dfrac14 = \dfrac{7}{12} in English and 1−14−13=5121 - \dfrac14 - \dfrac13 = \dfrac{5}{12} in History. Working over a denominator of 144:144: 88 points has probability 16⋅14=6144;\dfrac16\cdot\dfrac14 = \dfrac{6}{144}; 77 points has 16⋅13+14⋅14=17144;\dfrac16\cdot\dfrac13 + \dfrac14\cdot\dfrac14 = \dfrac{17}{144}; and 66 points has 16⋅512+14⋅13\dfrac16\cdot\dfrac{5}{12} + \dfrac14\cdot\dfrac13 +712⋅14=43144.+ \dfrac{7}{12}\cdot\dfrac14 = \dfrac{43}{144}. The total is 6+17+43144=66144=1124.\dfrac{6 + 17 + 43}{144} = \dfrac{66}{144} = \dfrac{11}{24}. Thus, the correct answer is D.
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