Question

Find the expected value of the winningsfrom a game that has the followingpayout probability distribution:Payout ($)12 5810Probability 0.35 0.2 0.1 0.2 0.15Expected Value = [?]

Answer 1

The expected value of the winnings is given by

$E(x)=\sum x\cdot p(x)$

Where x is the payout amount and p(x) is the corresponding probability.

$\begin{gathered} E(x)=\sum x\cdot p(x) \\ E(x)=(1\cdot0.35)+(2\cdot0.2)+(5\cdot0.1)+(8\cdot0.2)+(10\cdot0.15)_{} \\ E(x)=0.35+0.4+0.5+1.6+1.5 \\ E(x)=4.35 \end{gathered}$

This means that you are expected to get $4.35 in winnings.

Therefore, the expected value of the winnings is $4.35