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 A very bizarre weighted coin comes up heads with probability $\frac12$, tails with probability $\frac13$, and rests on its edge with probabi

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A very bizarre weighted coin comes up heads with probability $\frac12$, tails with probability $\frac13$, and rests on its edge with probability $\frac16$. If it comes up heads, I win 1 dollar. If it comes up tails, I win 3 dollars. But if it lands on its edge, I lose 5 dollars. What is the expected winnings from flipping this coin? Express your answer as a dollar value, rounded to the nearest cent.

 Apr 26, 2015

Best Answer 

 #1
avatar+128408 
+13

I believe the answer is this, Mellie

Expected Value =  (Amount won) x(Probability of winning) =

(1/2)($1) + (1/3)($3) + (1/6)(-$5)  =

.50 + 1.00 - .83 =

.67

 

  

 Apr 27, 2015

1+0 Answers

 #1
avatar+128408 
+13
Best Answer

I believe the answer is this, Mellie

Expected Value =  (Amount won) x(Probability of winning) =

(1/2)($1) + (1/3)($3) + (1/6)(-$5)  =

.50 + 1.00 - .83 =

.67

 

  

CPhill Apr 27, 2015

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