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

#1**+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**+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