What is a probability of coin being loaded?

Question

We have two coins: one is fair, and the other one is loaded with probability of $70\%$ coming head. Suppose we randomly chose a coin with prior probability of $60\%$ being a loaded one and flipped it $5$ times, and the result was $2$ heads $3$ tails. What is probability we chose a loaded coin?

Answer 1

I try to decrypt your attempt. So you have

$$P(\text{"loaded coin"}|H=2)=\frac{p(H=2| \text{"loaded coin"})\cdot P(\text{"loaded coin"})}{P(H=2)},$$

where $P(\text{"loaded coin"})=0.6$

$$ P(H=2| \text{"loaded coin"}) = 10\cdot 0.7^2\cdot 0.3^3$$

That´s correct. Additionally we have

$$ P(H=2| \text{"fair coin"}) = 10\cdot 0.5^2\cdot 0.5^3$$

Now we can use the Law of total probability: $P(H=2)$

$$=P(\text{"loaded coin"})\cdot P(H=2| \text{"loaded coin"})+P(\text{"fair coin"})\cdot P(H=2| \text{"fair coin"})$$

$=0.6\cdot 10\cdot 0.7^2\cdot 0.3^3+0.4\cdot 10\cdot 0.5^2\cdot 0.5^3$. Finally we get

$$P(\text{"loaded coin"}|H=2)=\frac{0.6\cdot 10\cdot 0.7^2\cdot 0.3^3}{0.6\cdot 10\cdot 0.7^2\cdot 0.3^3+0.4\cdot 10\cdot 0.5^2\cdot 0.5^3}=38.84\%$$