$$f(1)=\binom{4}{1}(\frac{1}{4})^{1}(\frac{3}{4})^{4-1}=$$
$\frac{27}{64}=0.421875$
$$4\times \left(\frac{1}{4}\right)^{1}\times \left(\frac{3}{4}\right)^{4-1}$$
$$4\times \frac{1}{4}\times \left(\frac{3}{4}\right)^{4-1}$$
$$\left(\frac{3}{4}\right)^{4-1}$$
$$\left(\frac{3}{4}\right)^{3}$$
$$\frac{27}{64}$$
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$\frac{3 ^ {3}}{2 ^ {6}} = 0.421875$
