Conditional Probability

1. Conditional Probability Terminology: The probability of an event occurring, given that another event has already occurred. : The probability of A given B.

2. Consider the following table: - The probability that a random selected smoker is male. - The probability that a random selected non-smoker is male. - The probability that a random selected smoker is female. - The probability that a random selected non-smoker is female. These probabilities can be represented in a tree diagram as shown on the right. It’s easy to see that: or

3. Now construct a conditional probability tree that includes the following probabilities: - The probability that a random selected male is a smoker. - The probability that a random selected male is a non-smoker. - The probability that a random selected female is a smoker. - The probability that a random selected female is a non-smoker. Answer on next slide! 

4. The probability that a random selected male is a smoker. The probability that a random selected male is a non-smoker. The probability that a random selected female is a smoker. The probability that a random selected female is a non-smoker.

5. What is the probability of a randomly selected individual being a male who smokes? • The number of "Male and Smoke" divided by the total = 19/100 = 0.19 • What is the probability of a randomly selected individual being a male? • This is the total for male divided by the total = 60/100 = 0.60. • What is the probability of a randomly selected individual smoking? • The total who smoke divided by the total = 31/100 = 0.31. • What is the probability of a randomly selected male smoking? • 19 males smoke out of 60 males, so 19/60 = 0.31666... • What is the probability that the male smokes? • There are 19 male smokers out of 31 total smokers, so 19/31 = 0.6129

6. Independent Events. Independent events have no effect on each other. For example, . Since, , we have The following are equivalent: The probability of either of two independent events or is: If , then which means that and are mutually exclusive events.

7. Now do Exercise 4B.  Also try Misc. Exercise 4

8. A bag contains 6 red and 4 green counters. Two counters are drawn, without replacement. Use a carefully labeled tree diagram to find the probabilities that Both counters are red Both counters are green Just one counter is red At least one counter is red The second counter is red Soln: OR b. OR c. OR

9. Soln: d. OR e. OR It makes no difference to (a), (b), (c) or (d) if the counters were drawn simultaneously, but it makes a difference to (e) because there would be no way to distinguish the second counter.

10. 2. Two cards are drawn, without replacement, from an ordinary pack. Find the probabilities that Both are picture cards Neither is a picture card At least one is a picture card At least one is red Soln: a. b. c. d.

11. 3. Events and satisfy these conditions: • Calculate • a. b. c. • Soln:

12. 4. A class consists of seven boys and nine girls. Two different members of the class are chosen at random. is the event , and is the event Find the probabilities of a. b. c. d. e. Is it true that f. ? g. ? Soln: a. b.

13. Soln: c. d. (*) Proved in (c). (*) Proof follows on next slide.

14. f. Is ? So identity is true. g. Is ? But , so Numerator Denominator, therefore

15. 5. A weather forecaster classifies all days as wet or dry. She estimates that the probability that 1 June next year is wet is 0.4. If any particular day in June is wet, the probability that the next day is wet is 0.6; otherwise the probability that the next day is wet is 0.3. Find the probability that next year, a. The first two days of June are both wet. b. June 2nd is wet. c. At least one of the first three days of June is wet. Soln: a. b. So, See tree diagram on next slide.

17. 6. Two chess players, K1 and K2, are playing each other in a series of games. The probability that K1 wins the first game is 0.3. If K1 wins any game, the probability that he wins the next is 0.4; otherwise the probability is 0.2. Find the probability that K1 wins The first two games At least one of the first two games The first three games Exactly one of the first three games The result of any game can be a win for K1, a win for K2, or a draw. The probability that any one game is drawn is 0.5, independent of the result of all previous games. Find the probability that after two games, e. K1 won the first and K2 the second f. Each won one game g. Each won the same number of games.

18. First we construct the conditional probability tree diagram. Note that on level 2, P(K1 W|W)=0.4 means the probability of K1 winning the second game, given that K1 has already won the first game (level 1), that is P(K1 W|W)=0.4

19. Soln: a. b. = So c. From the tree diagram, So

20. d. From the tree diagram, To solve (e), (f) and (g), we construct another tree diagram: e. f. g. We add 0.25 because all the previous games are equivalent to drawsor no-wins, since each player must have won the same number of games.

21. A fair cubical die is thrown four times. Find the probability that All four scores are 4 or more At least one score is less than 4 At least one of the scores is a 6 Soln: a. We only need concern ourselves with scores 4, 5 and 6. Since there are 3 such scores, and we are throwing the die four times, the number of desirable outcomes is . The sample space is given by . Therefore, the probability that all four scores are 4 or more is given by . b. c. To find the probability that at least one score is a 6, it is easier to find how many throws do not have a score of 6. It is evident this number is . Therefore the number of throws that have a score of 6 must be And so ,

22. 8. The Chevalier du Mere’s Problem. A seventeenth-century French gambler, the Chevalier du Mere, had run out of takers for his bet that, when a fair cubical dice was thrown four times, at least one 6 would be scored. (See Question 7c.). He therefore changed the game to throwing a pair of fair dice 24 times. What is the probability that, out of these 24 throws, at least one is a double 6? Soln: One double throw has equiprobable out comes of which are unfavorable to the bet. In throws, there are  possible outcomes of which only () are favorable. Thus the probability of that out of these throws, at least one is a double , is given by:

23. 9. The Birthday Problem. What is the probability that, out of randomly chosen people, at least two share a birthday? Assume that all 365 days of the year are equally likely and ignore leap years. (Hint: find the probabilities that two people have different birthdays, that three people have different birthdays, and so on.) Soln: IfP(A) is the probability of at least two people in the room having the same birthday, it may be simpler to calculate P(A'), the probability of there not being any two people having the same birthday. Then, because A and A' are the only two possibilities and are also mutually exclusive,P(A) = 1 − P(A'). If there are 23 randomly chosen people, the probability that there birthdays all fall on different days of the year is given by: OR Note that can be written as:

24. And so, Whence, In general, if there are people in a class room, then 10. Given that , and , calculate and . Soln: Now, So, But

25. 11. For any events and , write and in terms of , , and . Deduce Bayes’ theorem: Soln: We know that Now, and But,

26. 12. The Doctor’s Dilemma: It is known that among all patients displaying a certain set of symptoms, the probability that they have a particular rare disease is . A test for the disease has been developed. The test shows a positive result on of the patients who havew the disease and on of patients who do not have the disease. The test is given to a particular patient displaying the symptoms, and it records a positive result. Find the probability that the patient has the disease. Comment on your answer. Soln: Let be the event “rare disease” and be the event “positive result”. , so . You are given that and The probability that the patient has the disease is given by . Since the probability is very small, the test needs to be far more reliable so one can conclude there is sufficient evidence about a rare disease.

27. 13. The Prosecutor’s Fallacy: An accused prisoner is on trial. The defense lawyer asserts that in the absence of further evidence, the probability that the prisoner is guilty is 1 in a million. The prosecuting lawyer produces a further piece of evidence and asserts that if the prisoner were guilty, the probability that this evidence would be obtained is 999 in 1000, and if he were not guilty, it would only be 1 in 1000; in other words, , and . Assuming that the court admits the legality of the evidence, and that both lawyers’ figures are correct, what is the probability that the prisoner is guilty? Comment on your answer. Soln: Let be the event “guilty” and be the event “evidence”. , so . You are given that 99 and The probability that the prisoner is guilty is given by . The extra piece of evidence is not sufficient to make the prisoner’s guilt more likely; it would have to be far more certain.