# Conditional Probablity

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

## 1 Answer

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 $\mathbb{P}(A|B) = \frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)} \approx 0.85.$

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