Question

A game has 3 possible outcomes:The first outcome has a probability of 0.2. The result is winning $30.The second outcome has a probability of 0.2. The result is winning $50.The third outcome has a probability of 0.6. The result is losing $80.Let X be the amount of your winnings. Then E(X) =

Answer 1

We compute the expected value by multiplying each outcome by the probability of that outcome, next we add up those products.

$E(X)=\text{ 30}\cdot0.2\text{ + 50}\cdot0.2-80\cdot0.6=\text{ 6+10-48=-32}$