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

(Created page with "== Problem == ===Problem 19=== One day Jason finishes his math homework early, and decides to take a jog through his neighborhood. While jogging, Jason trips over a leprechaun. ...")
 
(Solution to Problem 19)
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== Problem ==
 
== Problem ==
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===Problem 19===
 
  
 
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!"
 
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!"
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\textbf{(E) }\dfrac14\qquad
 
\textbf{(E) }\dfrac14\qquad
 
\textbf{(F) }\dfrac13\qquad
 
\textbf{(F) }\dfrac13\qquad
−
\textbf{(G) }\dfrac25\qquad \\ </math>
+
\textbf{(G) }\dfrac25\qquad</math>
 +
 
 
<math>\textbf{(H) }\dfrac12\qquad
 
<math>\textbf{(H) }\dfrac12\qquad
 
\textbf{(I) }\dfrac35\qquad
 
\textbf{(I) }\dfrac35\qquad
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\textbf{(K) }\dfrac45\qquad
 
\textbf{(K) }\dfrac45\qquad
 
\textbf{(L) }1\qquad
 
\textbf{(L) }1\qquad
−
\textbf{(M) }\dfrac54\qquad \\ </math>
+
\textbf{(M) }\dfrac54\qquad</math>
 +
 
 
<math>\textbf{(N) }\dfrac43\qquad
 
<math>\textbf{(N) }\dfrac43\qquad
 
\textbf{(O) }\dfrac32\qquad
 
\textbf{(O) }\dfrac32\qquad
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== Solution ==
 
== Solution ==
 +
 +
The probability of getting <math>1</math> head is <math>\frac{1}{2}</math>.  The probability of getting <math>1</math> tail then <math>1</math> head is <math>\frac{1}{4}</math>.  From there, the probability of getting <math>n</math> tails then <math>1</math> head is <math>\frac{1}{2^{n+1}}</math>.
 +
 +
Let <math>E</math> be expected value of earnings.  Using expected value,
 +
<cmath>E = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + 2 \cdot \frac{1}{8} + 3 \cdot \frac{1}{16} \cdots</cmath>
 +
Multiply both sides by <math>2</math> to get
 +
<cmath>2E = 0 \cdot 1 + 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{8} \cdots</cmath>
 +
Subtract original to get
 +
<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>.
 +
 +
==See Also==
 +
{{iTest box|year=2007|num-b=18|num-a=20}}
 +
 +
[[Category:Intermediate Combinatorics Problems]]

Revision as of 16:07, 10 June 2018

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}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 18 		Followed by:
Problem 20
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