# Biased and fair coin in Hat flipped

Two coins are in a hat. The coins look alike, but one coin is fair (with probability 1/2 of Heads), while the other coin is biased, with probability 1/4 of Heads. One of the coins is randomly pulled from the hat, without knowing which of the two it is. Call the chosen coin “Coin A”.

Find the probability that in 10 flips of Coin A, there will be exactly 3 Heads.

PROPOSED SOLUTION:

P(H) = P(H|Fair)*P(Fair) + P(H|Biased)*P(Biased) = 1/2*1/2 + 1/8*1/2 = 3/8

P(T) = 1 - P(H) = 1 - 3/8 = 5/8

Now, P(3H in 10 flips) = (10C3)((3/8)^3)((5/8)^7) = 0.236

However the answer is given as 0.184.

Can anybody help understand the error in above approach. Thanks

If the difference is still not clear, first consider the easier case of $2$ tosses instead of $10$.
$P( \texttt{3 Heads}) = P(\texttt{3 Heads|fair})P(\texttt{fair}) + P(\texttt{3 Heads|biased})P(\texttt{biased})$
$=1/2\cdot{10\choose 3}\cdot(1/2)^3\cdot(1/2)^7 + 1/2\cdot{10\choose 3}\cdot (1/4)^3\cdot (3/4)^7$
$=0.184$