Difference between revisions of "2015 AMC 12B Problems/Problem 15"
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==Problem== | ==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 <math>\tfrac{1}{6}</math> chance of getting an A in English, and a <math>\tfrac{1}{4}</math> chance of getting a B. In History, she has a <math>\tfrac{1}{4}</math> chance of getting an A, and a <math>\tfrac{1}{3}</math> 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? | ||
| − | + | <math>\textbf{(A)}\; ? \qquad\textbf{(B)}\; ? \qquad\textbf{(C)}\; ? \qquad\textbf{(D)}\; ? \qquad\textbf{(E)}\; ?</math> | |
==Solution== | ==Solution== | ||
Revision as of 13:25, 3 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 
chance of getting an A in English, and a 
chance of getting a B. In History, she has a chance of getting an A, and a 
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?
Solution
See Also
| 2015 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
