Question

Find the expected value of the winninfrom a game that has the followingpayout probability distribution:Payout ($) 146 810Probability 0.12 0.2 0.38 0.2 0.1Expected Value = [?]Round to the nearest hundredth.

Answer 1

The expected value formula is given by;

`\text{Expected value= }\sum ^{}_{}xPr(x)`

Where x is the possible winning and Pr(x) is the probability of that same win.

Going through the table; we can compute the expected value of the winnings as;

`\text{Expected value=1(0.12)+4(0.2)}+6(0.38)+8(0.2)+10(0.1)=5.8`

Therefore, the expected value of winning is $5.8