Coin toss puzzle with 1 biased and 2 unbiased coins

Three coins are given: one two-head coin and two fair coins. You randomly choose a coin and the first three tosses give heads. What is the probability that 4-th toss is a head.

I have two solutions that give different results. Please help me to find the error.

1st solution:
$$P(\text{first 3 tosses are heads})=\frac{2}{3}\times\frac{1}{8}+\frac{1}{3}\times 1=\frac{5}{12}$$

$$P(\text{first 4 tosses are heads})=\frac{2}{3}\times\frac{1}{16}+\frac{1}{3}\times 1=\frac{3}{8}$$

$$P(\text{4th is head} \mid \text{first 3 tosses are heads})=\frac{P(\text{first 4 tosses are heads})}{P(\text{first 3 tosses are heads})}=\frac{9}{10}.$$

2nd solution:
$$P(\text{4th is head} \mid \text{first 3 tosses are heads})=$$
$$=\frac{2}{3}\times P(\text{4th is head} \mid \text{first 3 tosses are heads and coin is fair})+$$
$${}+\frac{1}{3}\times P(\text{4th is head}\mid\text{first 3 tosses are heads and coin is two-head})=\frac{2}{3}\times \frac{1}{2}+\frac{1}{3}\times 1=\frac{2}{3}.$$