# 1992 AHSME Problems/Problem 29

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

## 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 byProblem 28 Followed byProblem 30 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 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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