There are 15 coins in a bag. 5 of the 15 are fair coins and the rest are biased (80% H, 20% T). When a coin is chosen randomly from the bag and flipped twice, what is the probability that both of them are heads?

I tried to solve this using two different ways and I get two different answers.

Method 1:

$$P_{fair}(HH) = frac {1}{2^2}$$

Considering 80% is $frac 45$ and 20% is $frac 15$

$$P_{biased}(HH) = frac {4^2}{5^2}$$

$$P(HH) = frac {1}{3}P_{fair}(HH) + frac {2}{3}P_{biased}(HH)$$

$$P(HH) = frac {51}{100}$$

Method 2:

Constructing from all possible outcomes, consider 1 fair coin for every 2 biased coins.

$$P(HH) = frac {text{number of outcomes with two H}}{text{total outcomes}}$$

$$text{total outcomes} = (text{outcomes for fair coin}) + 2(text{outcomes for biased coin})$$

$$text{outcomes for fair coin} = text{{H, T} tossed twice}$$

$$text{outcomes for biased coin} = text{{H, H, H, H, T} tossed twice}$$

$$text{total outcomes}=2^2+2(5^2)$$

$$=54$$

$$text{number of outcomes with two H} = text{HH for fair coin} + 2(text{HH for biased coin})$$

$$=1+2(4^2)$$

$$=33$$

$$P(HH) = frac{33}{54}$$

Have I done a mistake in either of the methods or maybe both? It’s not like I didn’t understand conditional probability (or did I?). For instance, I can find the probability of drawing two red cards from a deck of playing cards using both those methods.

$$P(RR) = P(Red_1) P(Red_2|Red_1)$$

$$=frac{26}{52} frac{25}{51}$$

Also using just combinatorics,

$$P(RR) = frac{^{26}P_2}{^{52}P_2}$$

$$=frac{26 times 25}{52times51}$$

So definitely my methods aren’t incorrect. Coming back to my original question, where did I go wrong?