$\displaystyle \\\text{Given }Pr(X=i,\text{ }Y=j)=\frac{i+j-1}{8}\text{ for }i=1,\text{ }2\text{ and }j=1,\text{ }2.\\\\\text{Find }Var[Y].$

I know how to find the solution through finding $\displaystyle E[Y]$ and $\displaystyle E[Y^2]$, but my book gives an alternative solution that I haven't seen before.

Alternative solution:

$\displaystyle \\\text{Note:}\\\\\text{Alternatively, since }Y\text{ can take only two values, 1 and 2, with probabilities }\frac{3}{8}\text{ and }\frac{5}{8}\text{ respectively, we can use the Bernoulli shortcut to find }Var[Y].\\\\Var[Y]=(1-2)^2\left(\frac{3}{8}\right) \left(\frac{5}{8}\right)=0.2344$

I know that the variance of the Bernoulli distribution is $\displaystyle p(1-p)$, but if that's what the book is referring to, what's with the $\displaystyle (1-2)^2$ part?