# What is a probability of coin being loaded? [closed]

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?

• Do you know Bayes' rule? It is the tool to be used in this problem. – астон вілла олоф мэллбэрг May 5 '18 at 10:42
• i know but my calculation result is 0.4448 but the answer seems to be 0.388 – arash moradi May 5 '18 at 12:28
• Ah, ok. Now what you need to do is show your working. This way, we will be able to point out where you made a mistake. This is clearly more important than giving you a straight answer, since the correction of your thought process will leave you in a better position than merely showing the answer. – астон вілла олоф мэллбэрг May 5 '18 at 14:01
• It's still not clear. Please use the MathJax reference page to format your equations. I don't want to get your statements wrong, that's why I'm asking for clarity. If you do so, I shall upvote this question as well in appreciation of your effort. – астон вілла олоф мэллбэрг May 5 '18 at 16:08

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

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

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