Difference between revisions of "2015 AMC 12B Problems/Problem 17"
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| − | When tossed <math>n</math> times, the probability of getting exactly 2 heads and the rest tails could be written as <math>{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}</math>. However, we | + | When tossed <math>n</math> times, the probability of getting exactly 2 heads and the rest tails could be written as <math>{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}</math>. However, we must account for the different orders that this could occur in (for example like HTT or THT or TTH). We can account for this by multiplying by <math>\dbinom{n}{2}</math>. |
Similarly, the probability of getting exactly 3 heads is <math>\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}</math>. | Similarly, the probability of getting exactly 3 heads is <math>\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}</math>. | ||
Revision as of 13:12, 5 March 2015
Problem
An unfair coin lands on heads with a probability of 
. When tossed 
times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of ?
Solution
When tossed times, the probability of getting exactly 2 heads and the rest tails could be written as 
. However, we must account for the different orders that this could occur in (for example like HTT or THT or TTH). We can account for this by multiplying by 
.
Similarly, the probability of getting exactly 3 heads is 
.
Now set the two probabilities equal to each other and solve for :
![\[\dbinom{n}{2}{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}=\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}\]](http://latex.artofproblemsolving.com/d/e/e/deed9cb1861f787abde4c577294d240a24de5890.png)
![\[\frac{\dbinom{n}{2}}{\dbinom{n}{3}} = \frac{{\left(\frac{1}{4}\right)}^3 {\left(\frac{3}{4}\right)}^{n-3}}{{\left(\frac{1}{4}\right)}^2 {\left(\frac{3}{4}\right)}^{n-2}}\]](http://latex.artofproblemsolving.com/6/3/d/63db7df3a5ba5af6433cca6b839ab7ef536ef857.png)
![\[\frac{n(n-1)}{2!} \cdot \frac{3!}{n(n-1)(n-2)} = \frac{1}{4} \cdot {\left( \frac{3}{4} \right)}^{-1}\]](http://latex.artofproblemsolving.com/c/3/2/c32f68e7da37eb1c7b5f0194cd9261e206e28679.png)
![\[\frac{3}{n-2} = \frac{1}{3}\]](http://latex.artofproblemsolving.com/8/b/1/8b1cf0012808671f7758e6c3e60d5be2d0eaa713.png)
![\[n-2 = 9\]](http://latex.artofproblemsolving.com/1/c/d/1cd6ebfe15d5b338b44028394dc9b9543fe488f7.png)
![\[n = \fbox{\textbf{(D)}\; 11}\]](http://latex.artofproblemsolving.com/c/2/7/c27f5da250dd157f54a2a7e3c2e4eb2eb7e4ed48.png)
See Also
| 2015 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
