The Four Chocolate Coins

Relevant wiki: Probability - Rule of Product

The probability that any given coin is not eaten i.e. (tail, tail ) is (1/2) (1/2)=1/4.
The probability that a coin is eaten is (1-1/4)=3/4.
The probability of eating 4 coins is (3/4)

(3/4) (3/4) (3/4)= 81/256.
The probability of eating 3 coins is (3/4) (3/4) (3/4) (1/4) 4=108/256.
The probability of eating more than 2 coins is 81/256 + 108/256 = 189/256.
189+256=445

Let $X$ be the number of coins eaten at the first set of tosses, and let $Y$ be the number of coins eaten at the second set of tosses, so that $Z = X+Y$ is the total number of coins eaten. Then $\begin{array}{rcl} \mathbb{P}[Z \le 2] & = & \displaystyle \sum_{j=0}^2 \mathbb{P}[(Z \le 2) \cap (X = j)] \\ & = & \displaystyle \sum_{j=0}^2 \mathbb{P}[X=j]\,\mathbb{P}[Z \le 2 | X = j] \; = \; \sum_{j=0}^2 \mathbb{P}[X=j]\,\mathbb{P}[Y \le 2-j|X = j] \\ & = & \displaystyle \frac{1}{16}\times\frac{1+4+6}{16} + \frac{4}{16}\times\frac{1+3}{8} + \frac{6}{16}\times\frac{1}{4} \; = \; \frac{67}{256} \end{array}$ so that $\mathbb{P}[Z > 2] \,=\,1 - \tfrac{67}{256} \,=\, \tfrac{189}{256}$ , making the answer $\boxed{445}$ .