Difference between revisions of "2000 AMC 8 Problems/Problem 21"

Revision as of 20:35, 30 July 2011

Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is

$\text{(A)}\ \frac{1}{4}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{2}{3}\qquad\text{(E)}\ \frac{3}{4}$

Let $K(n)$ be the probability that Keiko gets $n$ heads, and let $E(n)$ be the probability that Ephriam gets $n$ heads.

$K(0) = \frac{1}{2}$

$K(1) = \frac{1}{2}$

$K(2) = 0$ (Keiko only has one penny!)

$E(0) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$

$E(1) = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} = 2\cdot\frac{1}{4} = \frac{1}{2}$ (because Ephraim can get HT or TH)

$E(2) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$

The probability that Keiko gets $0$ heads and Ephriam gets $0$ heads is $K(0)\cdot E(0)$ . Simiarly for $1$ head and $2$ heads. Thus, we have:

$P = K(0)\cdot E(0) + K(1)\cdot E(1) + K(2)\cdot E(2)$

$P = \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2} + 0$

$P = \frac{3}{8}$

Thus the answer is $\boxed{B}$ .

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

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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
 <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}}