### Instruction: 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.### Response: I know that the expected value of a random variable is the sum of the possible outcomes multiplied by their probabilities, so I can use that formula to calculate the expected winnings.
There are three possible outcomes: heads, tails, and edge.
The corresponding winnings are 1, 3, and -5 dollars.
The corresponding probabilities are $\frac12$, $\frac13$, and $\frac16$.
So the expected value is $(1)(\frac12) + (3)(\frac13) + (-5)(\frac16)$, which I can simplify by finding a common denominator of 6.
This gives me $\frac{3}{6} + \frac{6}{6} + \frac{-5}{6} = \frac{4}{6} = \frac{2}{3}$.
Therefore, the expected winnings from flipping this coin are $\frac{2}{3}$ dollars, or $0.67$ dollars, rounded to the nearest cent.