A loaded 8-sided die is loaded so that the number 4 occurs 3/10 of the time while the other numbers occur with equal frequency. what is the expected value of this die

The probability of not-4 to occur is:

P (not 4) = 1-P (4) = 1-3/10=7/10

The probability of one of {1, 2, 3, 5, 6} to happen is

P (1) + P (2) + P (3) + P (5) + P (6) = P (not 4) = 7/10

since all non-4 numbers have an equal chance to occur:

P (1) = P (2) = P (3) = P (5) = P (6) = (7/10) / 5=7/50=0.14

Thus the expected value is:

1*P (1) + 2*P (2) + 3*P (3) + 5*P (5) + 6*P (6) + 4*P (4)

= (1+2+3+5+6) * 0.14+4*0.3=2.38+1.2=3.58

Answer: 3.58