Why do these different methods vary? I randomly choose 1 of 2 coins. Coin A is fair: It has probability $\frac{1}{2}$ of yielding either heads or tails. Coin B is biased: It has a probability of $\frac{1}{4}$ of yielding heads and a probability of $\frac{3}{4}$ of yielding tails. If I toss my chosen coin 10 times, what's the probability that I will get exactly 3 heads?
I tried 2 different methods to solve this problem. Strangely, they produced different results. Which of these is right, and why do they differ? I thought they did the same thing. One just does it through the mean probability of getting heads in general.
Let $X$ be the number of heads out of 10 tosses.
Method 1:
P(1 head out of 1 toss) = $\frac{1}{2}\frac{1}{2} + \frac{1}{2}\frac{1}{4} = \frac{3}{8}$
Hence, $P(X=3) = \binom{10}{3}(\frac{3}{8})^3(\frac{5}{8})^7 \approx 0.235$
Method 2:
$P(X=3) = P(X=3|\text{ fair coin})P(\text{fair coin}) + P(X=3|\text{ biased coin})P(\text{ biased coin})$ 
$P(X=3) = \binom{10}{3}(\frac{1}{2})^3(\frac{1}{2})^7(\frac{1}{2}) + \binom{10}{3}(\frac{1}{4})^3(\frac{3}{4})^7(\frac{1}{2}) \approx 0.184$

The first method is right if you randomly choose your coin ten times and each time throw it once before choosing anew.
The second method is right if you choose one of the coins and stick with your choice for all ten throws. 
