Difference between revisions of "2015 AMC 12B Problems/Problem 15"

Revision as of 13:56, 5 March 2015

Problem

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

$\textbf{(A)}\; \frac{11}{72} \qquad\textbf{(B)}\; \frac{1}{6} \qquad\textbf{(C)}\; \frac{3}{16} \qquad\textbf{(D)}\; \frac{11}{24} \qquad\textbf{(E)}\; ?$

Solution

Solution 1

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 14 or more. 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 can 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 $\frac{1}{24} + \frac{17}{144} + \frac{31}{144} + \frac{1}{12} = \frac{66}{144} = \boxed{\mathbf{(D)}\; \frac{11}{24}}$ .

Solution 2

We can break it up into three mutually exclusive cases: A in english, at least a C in history; B in english and at least a B in history; C in english and an A in history. This gives

$\frac{1}{6} \cdot 1 + \frac{1}{4} \cdot \left(\frac{1}{3} + \frac{1}{4}\right) + \left(1 - \frac{1}{6} - \frac{1}{4}\right) \cdot \frac{1}{4} = \boxed{\textbf{(D)}\; \frac{11}{24}}.$

@@ Line 26: / Line 26: @@
 ==Solution 2==
-We can break it up into three cases: A in english, at least a C in history; B in english and at least a B in history; C in english and an A in history. This gives <math>(\frac{1}{6})(1) + (\frac{1}{4})(\frac{1}{3} + \frac{1}{4}) + (1 - \frac{1}{6} - \frac{1}{4})(\frac{1}{4}) = \frac{11}{24}</math>.
+We can break it up into three mutually exclusive cases: A in english, at least a C in history; B in english and at least a B in history; C in english and an A in history. This gives
+<cmath>\frac{1}{6} \cdot 1 + \frac{1}{4} \cdot \left(\frac{1}{3} + \frac{1}{4}\right) + \left(1 - \frac{1}{6} - \frac{1}{4}\right) \cdot \frac{1}{4} = \boxed{\textbf{(D)}\; \frac{11}{24}}.</cmath>
 ==See Also==
 {{AMC12 box|year=2015|ab=B|num-a=16|num-b=14}}
 {{MAA Notice}}

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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