Aloaded 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

3.58