Difference between revisions of "2000 AMC 8 Problems/Problem 21"
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+ | ==Problem== | ||
+ | |||
Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is | Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is | ||
<math> \text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4} </math> | <math> \text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4} </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | Let <math>K(n)</math> be the probability that Keiko gets <math>n</math> heads, and let <math>E(n)</math> be the probability that Ephriam gets <math>n</math> heads. | ||
+ | |||
+ | <math>K(0) = \frac{1}{2}</math> | ||
+ | |||
+ | <math>K(1) = \frac{1}{2}</math> | ||
+ | |||
+ | <math>K(2) = 0</math> (Keiko only has one penny!) | ||
+ | |||
+ | <math>E(0) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}</math> | ||
+ | |||
+ | <math>E(1) = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} = 2\cdot\frac{1}{4} = \frac{1}{2}</math> (because Ephraim can get HT or TH) | ||
+ | |||
+ | <math>E(2) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}</math> | ||
+ | |||
+ | The probability that Keiko gets <math>0</math> heads and Ephriam gets <math>0</math> heads is <math>K(0)\cdot E(0)</math>. Simiarly for <math>1</math> head and <math>2</math> heads. Thus, we have: | ||
+ | |||
+ | <math>P = K(0)\cdot E(0) + K(1)\cdot E(1) + K(2)\cdot E(2)</math> | ||
+ | |||
+ | <math>P = \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2} + 0</math> | ||
+ | |||
+ | <math>P = \frac{3}{8}</math> | ||
+ | |||
+ | Thus the answer is <math>\boxed{B}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC8 box|year=2000|num-b=20|num-a=22}} |
Revision as of 21:35, 30 July 2011
Problem
Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is
Solution
Let be the probability that Keiko gets
heads, and let
be the probability that Ephriam gets
heads.
(Keiko only has one penny!)
(because Ephraim can get HT or TH)
The probability that Keiko gets heads and Ephriam gets
heads is
. Simiarly for
head and
heads. Thus, we have:
Thus the answer is .
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 AJHSME/AMC 8 Problems and Solutions |