Difference between revisions of "1992 AHSME Problems/Problem 29"
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== Solution == | == Solution == | ||
| − | <math>\fbox{D}</math> | + | 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 10:33, 1 August 2016
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
