1992 AHSME Problems/Problem 29
Problem
An unfair coin has a 
probability of turning up heads. If this coin is tossed 
times, what is the probability that the total number of heads is even?
Solution
Doing casework on the number of heads (0 heads, 2 heads, 4 heads...), we get the equation ![\[P=\left(\frac{1}{3} \right)^{50}+\binom{50}{2}\left(\frac{2}{3} \right)^{2}\left(\frac{1}{3} \right)^{48}+\dots+\left(\frac{2}{3} \right)^{50}\]](http://latex.artofproblemsolving.com/0/c/a/0ca159135e493bb75d2c0367b2814f5788a9d797.png)
This is essentially the expansion of 
but without the odd power terms. To get rid of the odd power terms in , we add 
and then divide by 
because the even power terms that were not canceled were expressed twice. Thus, we have ![\[P=\frac{1}{2}\cdot\left(\left(\frac{1}{3}+\frac{2}{3} \right)^{50}+\left(\frac{2}{3}-\frac{1}{3} \right)^{50} \right)\]](http://latex.artofproblemsolving.com/3/7/9/379537954cb5ab96afb48a4a6de1a20262d63c45.png)
Or ![\[\frac{1}{2}\left(1+\left(\frac{1}{3} \right)^{50} \right)\]](http://latex.artofproblemsolving.com/e/5/7/e5743356ef3cc6fecfb7c1c7e3dec5bae28cd2b2.png)
which is equivalent to answer choice 
.
See also
| 1992 AHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
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