Let X be a binomially distributed random variable with mean 4 and...

Mean $np = 4$

Variance $npq = \dfrac{4}{3}$

i.e. $n = 6, p = \dfrac{2}{3}, q = \dfrac{1}{3}$

Now $54P\left(X \le 2\right) = 54\left[P\left(X = 2\right) + P\left(X = 1\right) + P\left(X = 0\right)\right]$

$= 54\left(C2 6 {\left(\dfrac{2}{3}\right)}^{2}{\left(\dfrac{1}{3}\right)}^{4} + C1 6 {\left(\dfrac{2}{3}\right)}^{1}{\left(\dfrac{1}{3}\right)}^{5} + C0 6 {\left(\dfrac{2}{3}\right)}^{0}{\left(\dfrac{1}{3}\right)}^{6}\right)$

$= 54\left(15 \times \dfrac{4}{{3}^{6}} + 6 \times \dfrac{2}{{3}^{6}} + \dfrac{1}{{3}^{6}}\right) = 2 \times {3}^{3}\left(\dfrac{73}{{3}^{6}}\right) = \dfrac{146}{27}$