+1836 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.
+529 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!
+33661 $${\mathtt{0.6}}{\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}\left(-{\mathtt{1}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{0.2}}{\mathtt{\,\times\,}}\left(-{\mathtt{10}}\right) = {\frac{{\mathtt{1}}}{{\mathtt{5}}}} = {\mathtt{0.2}}$$
Expected win = $0.20
