Conditional Probability and Independence

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 Thisrule 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 R1 and R2 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 Ai 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(Bc) > 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 F1, F2, and F3 are pairwise disjoint and S=F1F2 F3. • Now, • If event E is known to have occurred, then we can update the probabilities that the events Fj (the hypotheses) will occur by using Bayes’ formula. P(Fj) is called the prior probability of Fj and the conditional probability P(Fj | E) is the posterior probability of Fj 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 Fc. See proof in textbook. What can you say about the independence of Ec and Fc? • 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 E1, E2, …, En is independent if for every subset • Example. Suppose we conceive of an experiment involving an infinite number of coin flips. Suppose Ei 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 En 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(En) = (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 1st trial, G be event that a 7 occurs on 1st trial, and H be event that neither 5 nor 7 occurs on 1st 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 ?