Solutions - 5100-07

(i):

$$P[X=2]=\left(\genfrac{}{}{0px}{}{4}{2}\right){(0.5)}^2{(0.5)}^2=0.375$$

(ii):

$$P[X=3]=\left(\genfrac{}{}{0px}{}{6}{3}\right){(0.5)}^3{(0.5)}^3=0.3125$$

So event (i) is more probable.