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Difference between revisions of "1992 AHSME Problems/Problem 29"

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== Solution ==
 
== Solution ==
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<math>\fbox{D}</math>
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Doing casework on the number of heads (0 heads, 2 heads, 4 heads...), we get the equation <cmath>P=\left(\frac{1}{3} \right)^{50}+\binom{50}{2}\left(\frac{2}{3} \right)^{2}\left(\frac{1}{3} \right)^{48}+\dots+\left(\frac{2}{3} \right)^{50}</cmath> This is essentially the expansion of <math>\left(\frac{2}{3}+\frac{1}{3} \right)^{50}</math> but without the odd power terms. To get rid of the odd power terms in <math>\left(\frac{2}{3}+\frac{1}{3} \right)^{50}</math>, we add <math>\left(\frac{2}{3}-\frac{1}{3} \right)^{50}</math> and then divide by <math>2</math> because the even power terms that were not canceled were expressed twice. Thus, we have <cmath>P=\frac{1}{2}\cdot\left(\left(\frac{1}{3}+\frac{2}{3} \right)^{50}+\left(\frac{2}{3}-\frac{1}{3} \right)^{50} \right)</cmath> Or <cmath>\frac{1}{2}\left(1+\left(\frac{1}{3} \right)^{50} \right)</cmath> which is equivalent to answer choice <math>\fbox{D}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 09:33, 1 August 2016

Problem

An "unfair" coin has a $2/3$ probability of turning up heads. If this coin is tossed $50$ times, what is the probability that the total number of heads is even?

$\text{(A) } 25(\frac{2}{3})^{50}\quad \text{(B) } \frac{1}{2}(1-\frac{1}{3^{50}})\quad \text{(C) } \frac{1}{2}\quad \text{(D) } \frac{1}{2}(1+\frac{1}{3^{50}})\quad \text{(E) } \frac{2}{3}$

Solution

Doing casework on the number of heads (0 heads, 2 heads, 4 heads...), we get the equation \[P=\left(\frac{1}{3} \right)^{50}+\binom{50}{2}\left(\frac{2}{3} \right)^{2}\left(\frac{1}{3} \right)^{48}+\dots+\left(\frac{2}{3} \right)^{50}\] This is essentially the expansion of $\left(\frac{2}{3}+\frac{1}{3} \right)^{50}$ but without the odd power terms. To get rid of the odd power terms in $\left(\frac{2}{3}+\frac{1}{3} \right)^{50}$, we add $\left(\frac{2}{3}-\frac{1}{3} \right)^{50}$ and then divide by $2$ because the even power terms that were not canceled were expressed twice. Thus, we have \[P=\frac{1}{2}\cdot\left(\left(\frac{1}{3}+\frac{2}{3} \right)^{50}+\left(\frac{2}{3}-\frac{1}{3} \right)^{50} \right)\] Or \[\frac{1}{2}\left(1+\left(\frac{1}{3} \right)^{50} \right)\] which is equivalent to answer choice $\fbox{D}$.

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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All AHSME Problems and Solutions

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