# 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:

$P(X=3) = P(X=3|\text{ fair coin})P(\text{fair coin}) + P(X=3|\text{ biased coin})P(\text{ biased coin})$
$P(X=3) = \binom{10}{3}(\frac{1}{2})^3(\frac{1}{2})^7(\frac{1}{2}) + \binom{10}{3}(\frac{1}{4})^3(\frac{3}{4})^7(\frac{1}{2}) \approx 0.184$

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