A 6-sided die is loaded in such a way where odd outcomes occur 2/9 of the time, and even outcomes happen the other 1/9 of the time. What is this die’s expected value?

Exact answer as a fraction is 10/3

Approximate answer in decimal form is 3.333

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Explanation:

Make a table as shown below. We have X as the result of the die, so X takes on values between 1 and 6. The second column is the probability P(X) column. If X is odd, then P(X) = 2/9. Otherwise, P(X) = 1/9

Then make a third column that multiplies the X and P(X) values. For instance, in row 1 we have X*P(X) = 1*1/9 = 1/9. Row 2 will have X*P(X) = 2*2/9. And so on.

Once you compute all the X*P(X) values, add up the results

EV = Expected value

EV = sum of the X*P(X) values

EV = 2/9 + 2/9 + 6/9 + 4/9 + 10/9 + 6/9

EV = (2+2+6+4+10+6)/9

EV = 30/9

EV = 10/3

EV = 3.333 approximately

(23*2)-12=x     x=34

step-by-step explanation: