(Updaing posterior probabilities. ) Recall the setting of problem 1 form homework 3. Suppose that a box contains a fair coin and a coin with a head on each side. Suppose also that one coin is selected at random and that, when it is tossed, a head is observed. Let B1 the event that the selected coin is fair, B2 the event that the coin has two heads and H1 the event that a head is obtained when the selected coin is tossed. Using Bayes theorem, in problem 1 of homework 3 your showed that P(BI H1) = 1/3. Here P(B1) = 1/2 is the prior probability that the selected coin is fair, while P(B1| H1) = 1/3 is the posterior probability that the selected coin is fair. The posterior probability takes into account the additional information that the outcome of the coin toss was observed to be heads. Now suppose that the same coin is tossed a second time and that another head is obtained; call H2 this event. Assume that Hi and H2 are conditionally independent given Bi and B2 (this means that the outcomes of two tosses of the same coin are independent). We are now interested in computing P(B1 H1 n H2), the new or updated posterior probability that the selected coin is fair given that both the first and the second toss resulted in heads. There are two equivalent ways of calculating P(Bi|Hi n H2), described below. i. Compute P(Bi H1 n H2) using directly the definition of Bayes' theorem. ii. Compute P(BilHin H2) sequentially in two steps: first compute P(Bi|H1), the posterior probability of Bi given H1. This conditional probability now serves as a new prior proba- bility for the next stage of the experiment, in which the same coin is tossed a second time. In order to compute the final posterior probability you have to use the conditional version of Bayes' theorem from part (a). In both cases, you will find that P(Bi|H1 n H2) = 1/5, which is smaller than P(B1| H1) = 1/3, which in turn is smaller than P(B1) = 1/2. Explain