Answer
E(x) = $\frac{8}{3}$
Work Step by Step
The probability of selecting red marbles $= \frac{4}{4+2} = \frac{2}{3}$
R = red, G = green
P(x=0) → GGGG →$(\frac{1}{3})^{4}$
P(x=1) → RGGG, GRGG, GGRG, GGGR = $4(\frac{2}{3})(\frac{1}{3})^{3}$
P(x=2) → RRGG, GGRR, RGGR, GRRG, GRGR, RGRG = $6(\frac{2}{3})^{2}(\frac{1}{3})^{2}$
P(x=3) → RRRG, GRRR, RGRR, RRGR = $4(\frac{2}{3})^{3}(\frac{1}{3})$
P(x=4) → RRRR = $(\frac{2}{3})^{4}$
E(x) = $0\times (\frac{1}{3})^{4}+1\times 4(\frac{2}{3})(\frac{1}{3})^{3}+2\times 6(\frac{2}{3})^{2}(\frac{1}{3})^{2} +3\times 4(\frac{2}{3})^{3}(\frac{1}{3}) +4\times (\frac{2}{3})^{4} = \frac{8}{3}$
