1. Conditional Probability and Independence If A and B are events in sample space S and P(B) > 0, then the conditional probability of A given B is denoted P(A|B) and Example. A coin is flipped twice. Let S = {(H,H),(H,T),(T,H),(T,T)} and assume all four outcomes are equally likely. Let A be the event that both flips land on heads and let B be the event that at least one flip lands on heads. Since A = {(H,H)} and B = {(H,H), (H,T), (T,H)}, we have

2. Reduced Sample Space When working with conditional probability P(A|B), it is often easier to treat B as the new sample space. Example. A coin is flipped twice (continued). Let S = {(H,H),(H,T),(T,H),(T,T)}, A = {(H,H)}, B = {(H,H), (H,T), (T,H)}. Now, think of B as the sample space, where all outcomes are equally likely. Clearly, P(A|B) = 1/3, which agrees with the calculation on the previous slide.

3. The Law of Multiplication If we multiply through the conditional probability of A given B by P(B), we obtain the law of multiplication This rule can be generalized (see the textbook). Problem. Let an urn contain 8 red balls and 4 white balls. We draw 2 balls from the urn without replacement. If we assume that at each draw each ball in the urn is equally likely to be chosen, what is the probability that both balls are red? Solution. Let R 1 and R 2 denote, resp., the events that the first and second balls are red. Using the multiplication rule, we have Of course, we could solve this problem by a direct count without the use of conditional probability.

4. Probability of no king on four draws without replacement Draw from an ordinary deck of 52 cards Let A i be the event that no king is drawn on the ith draw. This is the same result we previously obtained by counting.

5. Law of Total Probability Let B be an event with P(B) > 0 and P(B c ) > 0. Then for any event A, This law may also be generalized--see textbook. Example. An insurance company rents 35% of the cars for its customers from agency I and 65% from agency II. If 8% of the cars of agency I and 5% of the cars of agency II break down during the rental periods, what is the probability that a car rented by this insurance company breaks down? A tree diagram is often useful for the law of total probability.

6. Bayes’ Formula--see text for a generalization Suppose F 1, F 2, and F 3 are pairwise disjoint and S=F 1  F 2  F 3. Now, If event E is known to have occurred, then we can update the probabilities that the events F j (the hypotheses) will occur by using Bayes’ formula. P(F j ) is called the prior probability of F j and the conditional probability P(F j | E) is the posterior probability of F j after the occurrence of E.

7. Example for Bayes’ formula Suppose we have 3 cards which are identical in form. The first card has both sides red, the second card has both sides black, and the third card has one red side and one black side. Suppose that one of the cards is randomly selected and put down on the ground. If the upturned side of the chosen card is red, what is the probability that the other side is black? Let R2, B2, and M denote the events that the chosen card is, resp., all red, all black, and mixed (red-black). Letting R be the event that the upturned side of the chosen card is red, we have

8. Independent events In general, P(E|F) and P(E) are different. That is, knowing that F has occurred generally changes the probability of E’s occurrence. This leads to the following definition. Events E and F are independent in case If E and F are not independent, we say they are dependent. Example. Two coins are flipped and all 4 outcomes are assumed to be equally likely. If E is the event that the first coin lands heads and F is the event that the second coin lands tails, then E and F are independent since

9. More on independent events If E and F are independent, so are E and F c. See proof in textbook. What can you say about the independence of E c and F c ? Assuming P(EFG) = P(E)P(F)P(G) for three events E, F, G does not imply pairwise independence. See Example 3.29 in the textbook. We say E 1, E 2, …, E n is independent if for every subset Example. Suppose we conceive of an experiment involving an infinite number of coin flips. Suppose E i is the event that the ith flip turns up heads. We believe that these events are independent, and this means that all equations of type * will hold (without the restriction that subscripts are n). This shows how to extend the concept of independence to a sequence of events.

10. Example—An experiment with independent subexperiments Independent trials, consisting of rolling a pair of dice are performed. What is the probability of the event E that we get a sum of 5 before we get a sum of 7? Let E n denote the event that no 5 or 7 appears on the first n–1 trials and a 5 appears on the nth trial. The desired probability is Since P({5 on any trial}) = 4/36 and P({7 on any trial}) = 6/36, by the independence of trials, P(E n ) = (1– 10/36) n-1 (4/36) and thus the desired probability is 2/5 using the result of Appendix 2. Let F be event that a 5 occurs on 1 st trial, G be event that a 7 occurs on 1 st trial, and H be event that neither 5 nor 7 occurs on 1 st trial. P(E) = P(E|F)P(F)+P(E|G)P(G)+P(E|H)P(H). Now P(E|F) = 1 and P(E|G) = 0. P(E|H) = P(E) since H has no effect. We have, P(E) = 1/9 +P(E)(13/18) or P(E) = 2/5. This is the same result as before. Do you see why ?