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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.

Respuesta :

The expected value formula is given by;

[tex]\text{Expected value= }\sum ^{}_{}xPr(x)[/tex]

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;

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

Therefore, the expected value of winning is $5.8