Track / Stage 3 / 177 of 260 #177 of 2000

Problem 177

AMC 10/12, early questions

Combinatorics Difficulty 3.6 Find the answer

At Rachelle's school, an A counts $4$ points, a B $3$ points, a C $2$ points, and a D $1$ 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 $\frac{1}{6}$ chance of getting an A in English, and a $\tfrac{1}{4}$ 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 $3.5$ ?
$\textbf{(A)}\; \frac{11}{72} \qquad\textbf{(B)}\; \frac{1}{6} \qquad\textbf{(C)}\; \frac{3}{16} \qquad\textbf{(D)}\; \frac{11}{24} \qquad\textbf{(E)}\; \frac{1}{2}$

Your answer

Official solution

The probability that Rachelle gets a C in English is $1-\frac{1}{6}-\frac{1}{4} = \frac{7}{12}$ .
The probability that she gets a C in History is $1-\frac{1}{4}-\frac{1}{3} = \frac{5}{12}$ .
We see that the sum of Rachelle's "point" scores must be at least 14 since $4*3.5 = 14$ . 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 $\frac{1}{6}\cdot\frac{1}{4} = \frac{1}{24}$ .
The probability that she gets one A and one B is $\frac{1}{6}\cdot\frac{1}{3} + \frac{1}{4}\cdot\frac{1}{4} = \frac{1}{18}+\frac{1}{16} = \frac{8}{144}+\frac{9}{144} = \frac{17}{144}$ .
The probability that she gets one A and one C is $\frac{1}{6}\cdot\frac{5}{12} + \frac{1}{4}\cdot\frac{7}{12} = \frac{5}{72}+\frac{7}{48} = \frac{31}{144}$ .
The probability that she gets two Bs is $\frac{1}{4}\cdot\frac{1}{3} = \frac{1}{12}$ .
Adding these, we get $\frac{1}{24} + \frac{17}{144} + \frac{31}{144} + \frac{1}{12} = \frac{66}{144} = \boxed{\mathbf{(D)}\; \frac{11}{24}}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.