Maths Olympiad Prep

Track / Stage 3 / 202 of 260 #202 of 1964

Problem 202

AMC 10/12, early questions
Combinatorics Difficulty 3.6 Multiple choice

At Rachelle's school, an A counts 44 points, a B 33 points, a C 22 points, and a D 11 point. Her GPA in the four classes she is taking is computed as the total sum of points divided by 4. She is certain that she will get A's in both Mathematics and Science and at least a C in each of English and History. She thinks she has a 16\frac{1}{6} chance of getting an A in English, and a 14\tfrac{1}{4} chance of getting a B. In History, she has a 14\frac{1}{4} chance of getting an A, and a 13\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 3.53.5?

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Official solution

The probability that Rachelle gets a C in English is 11614=7121-\frac{1}{6}-\frac{1}{4} = \frac{7}{12}.
The probability that she gets a C in History is 11413=5121-\frac{1}{4}-\frac{1}{3} = \frac{5}{12}.
We see that the sum of Rachelle's "point" scores must be at least 14 since 43.5=144*3.5 = 14. We know that in Mathematics and Science we have a total point score of 8 (since she will get As in both), so we only need a sum of 6 in English and History. This can be achieved by getting two As, one A and one B, one A and one C, or two Bs. We evaluate these cases.
The probability that she gets two As is 1614=124\frac{1}{6}\cdot\frac{1}{4} = \frac{1}{24}.
The probability that she gets one A and one B is 1613+1414=118+116=8144+9144=17144\frac{1}{6}\cdot\frac{1}{3} + \frac{1}{4}\cdot\frac{1}{4} = \frac{1}{18}+\frac{1}{16} = \frac{8}{144}+\frac{9}{144} = \frac{17}{144}.
The probability that she gets one A and one C is 16512+14712=572+748=31144\frac{1}{6}\cdot\frac{5}{12} + \frac{1}{4}\cdot\frac{7}{12} = \frac{5}{72}+\frac{7}{48} = \frac{31}{144}.
The probability that she gets two Bs is 1413=112\frac{1}{4}\cdot\frac{1}{3} = \frac{1}{12}.
Adding these, we get 124+17144+31144+112=66144=(D)  1124\frac{1}{24} + \frac{17}{144} + \frac{31}{144} + \frac{1}{12} = \frac{66}{144} = \boxed{\mathbf{(D)}\; \frac{11}{24}}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.