Difference between revisions of "2015 AMC 12B Problems/Problem 17"

Revision as of 12:10, 11 January 2021

Problem

An unfair coin lands on heads with a probability of $\tfrac{1}{4}$ . When tossed $n>1$ times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of $n$ ?

$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\; 13$

Solution

When tossed $n$ times, the probability of getting exactly 2 heads and the rest tails is

$\dbinom{n}{2} {\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}.$

Similarly, the probability of getting exactly 3 heads is

$\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.$

Now set the two probabilities equal to each other and solve for $n$ :

$\dbinom{n}{2}{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}=\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}$

$\frac{n(n-1)}{2!} \cdot \frac{3}{4} = \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{4}$

$3 = \frac{n-2}{3}$

$n-2 = 9$

$n = \fbox{\textbf{(D)}\; 11}$

Note: the original problem did not specify $n>1$ , so $n=1$ was a solution, but this was fixed in the Wiki problem text so that the answer would make sense. — @adihaya (talk) 15:23, 19 February 2016 (EST)

Solution 2

Bash it out with the answer choices! (not really a rigorous solution)

Solution 2.5

In order to test the answer choices efficiently, realize that the probability $n$ flips yielding two heads is of the form:

$\dbinom{n}{2}{\left(\frac{1}{4} \cdot \frac{1}{4}\right)}{\left(\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \ldots \right)} = \dbinom{n}{2}{\left(\frac{3^{n-2}}{4^n}\right)}$

Similarly, the form for the probability of three heads is:

$\dbinom{n}{3}{\left(\frac{3^{n-3}}{4^n}\right)}$ .

The probability of getting three heads (comapred to the probability of getting two) from $n$ flips is missing a factor of three in the numerator. Thus, we need $\dbinom{n}{3}$ to add a factor of 3 to the numerator of the probability of getting three heads. Our testing equation becomes:

$\dbinom{n}{2} \times 3 = \dbinom{n}{3}$ .

Since after factoring out the three from $\dbinom{n}{3}$ , the remaining factorizations should be equal.

The only answer choice satisfying this condition is $\fbox{\textbf{(D)}\;11}$ .

-Solution by Joeya

@@ Line 31: / Line 31: @@
 Bash it out with the answer choices! (not really a rigorous solution)
+==Solution 2.5==
+In order to test the answer choices efficiently, realize that the probability <math>n</math> flips yielding two heads is of the form:
+<math>\dbinom{n}{2}{\left(\frac{1}{4} \cdot \frac{1}{4}\right)}{\left(\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \ldots \right)} = \dbinom{n}{2}{\left(\frac{3^{n-2}}{4^n}\right)}</math>
+Similarly, the form for the probability of three heads is:
+<math>\dbinom{n}{3}{\left(\frac{3^{n-3}}{4^n}\right)}</math>.
+The probability of getting three heads (comapred to the probability of getting two) from <math>n</math> flips is missing a factor of three in the numerator. Thus, we need <math>\dbinom{n}{3}</math> to add a factor of 3 to the numerator of the probability of getting three heads.
+Our testing equation becomes:
+<math>\dbinom{n}{2} \times 3 = \dbinom{n}{3}</math>.
+Since after factoring out the three from <math>\dbinom{n}{3}</math>, the remaining factorizations should be equal.
+The only answer choice satisfying this condition is <math>\fbox{\textbf{(D)}\;11}</math>.
+-Solution by Joeya
 ==See Also==
 {{AMC12 box|year=2015|ab=B|num-a=18|num-b=16}}
 {{MAA Notice}}

2015 AMC 12B (Problems • Answer Key • Resources)
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