100 coins with two sides (head and tail)
20 coins are fair (50% of getting head and 50% of getting tail)
80 coins are biased (70% of getting head and 30% of getting tail)

What is the probability of get head if we throw a randomly chosen coin from the 100 coins once I said .2*.5 + .8*.7 = .66

Now, given that we got a head, what is the conditional probability that the coin we threw was biased? This is tricky for me as to how to set it up. I understand that the formula is P(A|B) = P(A intersect B)/ P(B)

but im not sure how to get P(A intersect B) or P(B) or how to assign those


Let $A$ be the event that you the coin you chose is biased and $B$ the event that the coin shows heads. Then $A\cap B$ is the event that your chosen coin is biased and shows heads, i.e. $\mathbb{P}(A\cap B) = 0.8 * 0.7$ and, as you correctly calculated, $\mathbb{P}(B) = 0.66$. Therefore $\mathbb{P}(A|B) = \frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)} \approx 0.85.$

  • $\begingroup$ very clear! thank you $\endgroup$ – antz Nov 14 '12 at 3:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.