# Coin toss

• Aug 22nd 2009, 12:40 PM
morganfor
Coin toss
A girl picks between one of two coins (probability of choosing each is 1/2) and one of the coins is fair and the other has two heads. Given that she reports a head on the nth toss, what is the probability that a head is thrown on the (n+1)st toss?

They are getting 5/6 as an answer but I can't figure out how to get that
• Aug 22nd 2009, 01:12 PM
rn443
Quote:

Originally Posted by morganfor
A girl picks between one of two coins (probability of choosing each is 1/2) and one of the coins is fair and the other has two heads. Given that she reports a head on the nth toss, what is the probability that a head is thrown on the (n+1)st toss?

They are getting 5/6 as an answer but I can't figure out how to get that

Let F be the hypothesis that the coin is fair, B be the hypothesis that the coin is biased, E be the evidence that the coin just landed on heads, and N be the statement that the coin will land on heads next time. Then the question is asking us to evaluate $P(N|E)$. We can apply the axioms of probability theory and the definition of conditional probability to show that $P(N|E) = P(N|E\&F)P(F|E) + P(N|E\&B)P(B|E)$. See if you can figure out what the expressions P(N|E&F) and P(N|E&B) are. Once you do, the only thing remaining is to figure out P(F|E) and P(B|E).

To do this, apply Bayes' theorem. Bayes' theorem tells us that $P(X|Y) = \frac{P(Y|X)P(X)}{P(Y|X)P(X) + P(Y|\neg X)P(\neg X)}.$ Figure out what to plug in for each of the probabilities.