Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2007 iTest Problems/Problem 19"

(Solution to Problem 19)
m (→‎Solution)
Line 38: Line 38:
 
Subtract original to get
 
Subtract original to get
 
<cmath>E = 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + 1 \cdot \frac{1}{8} \cdots</cmath>
 
<cmath>E = 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + 1 \cdot \frac{1}{8} \cdots</cmath>
−
Using the infinite series formula, <math>E = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1</math>, so Jason is expected to earn <math>\boxed{\textbf{(L) } \$ 1}</math>.
+
Using the infinite series formula, <math>E = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1</math>, so Jason is expected to earn <math>\boxed{\textbf{(L) } \1 \text{gold coin}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 00:17, 29 October 2023

Problem

One day Jason finishes his math homework early, and decides to take a jog through his neighborhood. While jogging, Jason trips over a leprechaun. After dusting himself off and apologizing to the odd little magical creature, Jason, thinking there is nothing unusual about the situation, starts jogging again. Immediately the leprechaun calls out, "hey, stupid, this is your only chance to win gold from a leprechaun!" Jason, while not particularly greedy, recognizes the value of gold. Thinking about his limited college savings, Jason approaches the leprechaun and asks about the opportunity. The leprechaun hands Jason a fair coin and tells him to clop it as many times as it takes to flip a head. For each tail Jason flips, the leprechaun promises one gold coin. If Jason flips a head right away, he wins nothing. If he first flips a tail, then a head, he wins one gold coin. If he's lucky and flips ten tails before the first head, he wins $\textit{ten gold coins.}$ What is the expected number of gold coins Jason wins at this game?

$\textbf{(A) }0\qquad \textbf{(B) }\dfrac1{10}\qquad \textbf{(C) }\dfrac18\qquad \textbf{(D) }\dfrac15\qquad \textbf{(E) }\dfrac14\qquad \textbf{(F) }\dfrac13\qquad \textbf{(G) }\dfrac25\qquad$

$\textbf{(H) }\dfrac12\qquad \textbf{(I) }\dfrac35\qquad \textbf{(J) }\dfrac23\qquad \textbf{(K) }\dfrac45\qquad \textbf{(L) }1\qquad \textbf{(M) }\dfrac54\qquad$

$\textbf{(N) }\dfrac43\qquad \textbf{(O) }\dfrac32\qquad \textbf{(P) }2\qquad \textbf{(Q) }3\qquad \textbf{(R) }4\qquad \textbf{(S) }2007$


Solution

The probability of getting $1$ head is $\frac{1}{2}$. The probability of getting $1$ tail then $1$ head is $\frac{1}{4}$. From there, the probability of getting $n$ tails then $1$ head is $\frac{1}{2^{n+1}}$.

Let $E$ be expected value of earnings. Using expected value, \[E = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + 2 \cdot \frac{1}{8} + 3 \cdot \frac{1}{16} \cdots\] Multiply both sides by $2$ to get \[2E = 0 \cdot 1 + 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{8} \cdots\] Subtract original to get \[E = 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + 1 \cdot \frac{1}{8} \cdots\] Using the infinite series formula, $E = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$, so Jason is expected to earn $\boxed{\textbf{(L) } \1 \text{gold coin}}$ (Error compiling LaTeX. Unknown error_msg).

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 18 		Followed by:
Problem 20
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_19&oldid=200429"