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?

  • $\begingroup$ Do you know Bayes' rule? It is the tool to be used in this problem. $\endgroup$ – астон вілла олоф мэллбэрг May 5 '18 at 10:42
  • $\begingroup$ i know but my calculation result is 0.4448 but the answer seems to be 0.388 $\endgroup$ – arash moradi May 5 '18 at 12:28
  • $\begingroup$ 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. $\endgroup$ – астон вілла олоф мэллбэрг May 5 '18 at 14:01
  • $\begingroup$ p(loaded | H=2)=p(H=2| loaded)*.6/p(H=2) ,p(H=2)= 10*(.5)^5+10*(.7)^2*(.3)^3 ,p(H=2| loaded) = 10*(.7)^2*(.3)^3 $\endgroup$ – arash moradi May 5 '18 at 14:35
  • $\begingroup$ 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. $\endgroup$ – 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 $P(\text{"loaded coin"})=0.6$

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

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

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

  • 1
    $\begingroup$ Peculiar use of the word encrypt. $\endgroup$ – Gerry Myerson Oct 10 at 5:33
  • 1
    $\begingroup$ Oh, Oh, Oh. Good catch. I wanted to write decrypt. I hope it´s better!? $\endgroup$ – callculus Oct 10 at 5:39

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