Prove that $(\frac{13}{16})^{12}=\sum_{k=0}^{12}{12 \choose k}\left(\frac{1}{4}\right)^{2k}\left(\frac{3}{4}\right)^{12-k}$

Question

I tried to solve a probability question where I got $\sum_{k=0}^{12}{12 \choose k}\left(\frac{1}{4}\right)^{2k}\left(\frac{3}{4}\right)^{12-k}$ but did not succeed in evaluating that to $(\frac{13}{16})^{12}$ by hand. There is an answer that requires defining a specific probability space, but my question is whether there's a more "straightforward" way of doing this proof.

Answer 1

This is nothing but the binomial expansion.

$$\frac{13}{16} = \frac{1}{16} + \frac{12}{16}$$

hence

$$\left(\frac{13}{16}\right)^{12} = \left(\frac{1}{16} + \frac{12}{16}\right)^{12} = \sum_{k =0}^{12} \binom{12}{k}\left(\frac{1}{16}\right)^k \left(\frac{12}{16}\right)^{12-k}$$

Now just notice that

$$\frac{12}{16} = \frac{3}{4}$$ and $$\frac{1}{16} = \left(\frac{1}{4}\right)^2$$

whence the identity.

Binomial Expansion Review

$$(A + B)^n = \sum_{k =0}^n \binom{n}{k} A^n B^{n-k}$$