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 (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS