help me find what is expected number of green balls in the sample

(a)

Prob 0 green  =  5C3 / 9C3  = .12

Prob  1 green  =  3C1 * (4/9)(5/9)^2  = .41

Prob 2 green 3C2 * (4/9)^2 * (5/9)  = .33

Prob 3 green  = (4/9)^3  = .09

The last histogram is correct

(b)

Expected number of green balls

0*(.12) + 1*(.41)  + 2*(.33) + 3*(.09)  = 1.34