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An unfair coin lands on heads with probability $\frac35$, on tails with probability $\frac15$, and on its edge with probability $\frac15$. If it comes up heads, I win 4 dollars. If it comes up tails, I lose 1 dollar. But if it lands on its edge, I lose 10 dollars. What is the expected winnings from one flip? Express your answer as a dollar value, rounded to the nearest cent.

Mellie  Apr 28, 2015

Best Answer 

 #1
avatar+520 
+15

Hi Mellie.

expected value,  E(x) = 0.6*4 + 0.2*-1 + 0.2*-10 

= 2.18

≈ $2

 

Probabilty carries no certainty. Knowing my luck, if I was given 10 throws, I'd come out $100 poorer!

 

  

Badinage  Apr 28, 2015
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2+0 Answers

 #1
avatar+520 
+15
Best Answer

Hi Mellie.

expected value,  E(x) = 0.6*4 + 0.2*-1 + 0.2*-10 

= 2.18

≈ $2

 

Probabilty carries no certainty. Knowing my luck, if I was given 10 throws, I'd come out $100 poorer!

 

  

Badinage  Apr 28, 2015
 #2
avatar+26399 
+10

$${\mathtt{0.6}}{\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}\left(-{\mathtt{1}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}\left(-{\mathtt{10}}\right) = {\frac{{\mathtt{1}}}{{\mathtt{5}}}} = {\mathtt{0.2}}$$

 Expected win = $0.20 

Alan  Apr 29, 2015

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