Conditional Probability

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

2 Consider the following table: YES NO TOTAL MALE FEMALE P M S = P M NS = P F S = P F NS = - 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 P A and B = P A P B A P A B = P A P B A

3 Now construct a conditional probability tree that includes the following probabilities: P S M P NS M P S F P F NS - 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. P S M = The probability that a random selected male is a non-smoker. P NS M = The probability that a random selected female is a smoker. P S F = The probability that a random selected female is a non-smoker. P NS F = YES NO TOTAL MALE FEMALE

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 = What is the probability of a randomly selected individual smoking? The total who smoke divided by the total = 31/100 = What is the probability of a randomly selected male smoking? 19 males smoke out of 60 males, so 19/60 = What is the probability that the male smokes? There are 19 male smokers out of 31 total smokers, so 19/31 = YES NO TOTAL MALE FEMALE

6 Independent Events. Independent events have no effect on each other. For example, P A B = P(A). Since, P A B = ( ) (), we have P A = P A and B = P(A) P(B) The following are equivalent: P A and B = P(A) P(B) P A B = P(A) P(B) The probability of either of two independent events A or B is: P A B = P A + P B P(A B) If P A B = 0, then P A B = P A + P B which means that A and B are mutually exclusive events.

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

8 1. 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 a. Both counters are red b. Both counters are green c. Just one counter is red d. At least one counter is red e. The second counter is red Soln: a. P R and R = = OR = b. P G and G = = OR = c. P R or R = = OR + = + =

9 Soln: d. P at least one R = = OR 1 = e. P second counter R = = OR = = = P second counter R = P at least one R P R followed by G 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 a. Both are picture cards (K, Q, J) b. Neither is a picture card c. At least one is a picture card d. At least one is red Soln: a. P Both picture cards = = b. P Neither is a picture card = = = c. P At least one is a picture card = 1 P Neither is a picture card = 1 = d. P At least one is red = 1 P Both are black = 1 = 1 = 1 =

11 3. Events A, B and C satisfy these conditions: P A = 0.6 P B = 0.8 P B A = 0.45 P B and C = 0.28 Calculate a. P(A and B) b. P(C B) c. P(A B) Soln: a. P A and B = P(B A) P(A) = = 0.27 b. ( ) P C B = =. (). c. ( ) P A B = =. ().

12 4. A class consists of seven boys and nine girls. Two different members of the class are chosen at random. A is the event {first person is a girl}, and B is the event the second person is a girl. Find the probabilities of a. B A b. B A c. B A d. B A e. B Is it true that f. P B A + P B A = 1? g. P B A + P B A = 1? Soln: a. P B A = ( ) () b. P B A = ( ) () = = = ( ) () = = =

13 Soln: c. P B = + = = P B A = ( ) ( ) = ( ) () = = d. P B A = 1 P B A = 1 = (*) e. Proved in (c). (*) Proof follows on next slide.

14 f. Is P B A + P B A = 1? ( ) () = ()( ) () = 1 So identity is true. g. Is P B A + P B A = 1? () + ( ) () = () () () () = ( ) ()() () () But P A and B = P B P B A, so = ( ) () () () = ( )() () () Numerator Denominator, therefore P B A + P B A 1

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 2 nd is wet. c. At least one of the first three days of June is wet. Soln: a. P 1st and 2nd wet = = 0.24 b. P 2nd wet = = = 0.42 c. P At least 1 wet = 1 P None wet P None wet = P first day dry P 2nd dry 1st dry) P(3rd dry 2nd dry) P None wet = = So, P At least 1 wet = = 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 a. The first two games b. At least one of the first two games c. The first three games d. 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. P K1 wins first two games = = 0.12 b. P K1 wins at least one of the first two games = 1 P(K1 wins neither of the first two games) P(K1 wins neither of the first two games)= P K1 loses first game P K1 loses second game K1 lost 1st game) P K1 wins neither of the first two games = = 0.56 So P K1 wins at least one of the first two games = = 0.44 c. From the tree diagram, P(K1 wins the first three games) = P K1 wins first game P K1 wins 2nd game K1 won 1st game) P K1 wins 3rd game K1 won 2nd game) So P(K1 wins the first three games) = = 0.048

20 d. From the tree diagram, P K1 wins exactly one of the first three games = P K1 loses two games = = To solve (e), (f) and (g), we construct another tree diagram: e. P K1 won first and K2 second = 0.03 ( ) f. P each won one game = = g. P(each has won the same number of games) = We add 0.25 because all the previous games are equivalent to draws or no-wins, since each player must have won the same number of games.

21 7. A fair cubical die is thrown four times. Find the probability that a. All four scores are 4 or more b. At least one score is less than 4 c. 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 3 = 81. The sample space is given by 6 = Therefore, the probability that all four scores are 4 or more is given by =. b. P At least one score is less than 4 = 1 P All four scores are 4 or more P At least one score is less than 4 = 1 = 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 5 = 625. Therefore the number of throws that have a score of 6 must be 6 5 = = 671. And so, P At least one of the scores is a 6 =

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 36 equiprobable out comes of which 35 are unfavorable to the bet. In 24 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 6, is given by: =... =

23 9. The Birthday Problem. What is the probability that, out of 23 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: If P(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: P A = OR P A = Note that can be written as:!! = !

24 And so, P A = ! Whence, P A = ! = = In general, if there are n people in a class room, then P A = n n! 10. Given that P A = 0.75, P B A = 0.8 and P B A = 0.6, calculate P(B) and P(A B). Soln: P B A = ( ) ( ) = 0.6 ( ) () = 0.6 Now, P B A = ( ) () = 0.8 ( ). = 0.8 P A and B = = 0.6 So, ( ) () = = 0.6 P B = 0.75 But P A B = ( ) () =.. = 0.8

25 11. For any events A and B, write P(A and B) and P(B) in terms of P(A), P(A ), P(B A) and P(B A ). Deduce Bayes theorem: Soln: P A B = ( ) ( ) We know that P A and B = P B P(A and B) P(B) = P A and B + P(A and B) Now, P A and B = P A P(B A ) P(A and B) = P A P B A and But, P A B = ( ) () = () ( ) = () ( )

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 98% of the patients who havew the disease and on 3% 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 A be the event rare disease and B be the event positive result. P A = 0.001, so P A = You are given that P B A = 0.98 and P B A = 0.03 The probability that the patient has the disease is given by P(A B). P A B = (.)(.).. (.)(.) = 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, P E G = 0.999, and P E G = 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 G be the event guilty and E be the event evidence. P G = , so P G = You are given that P E G = and P E G = The probability that the prisoner is guilty is given by P(G E). P G E = (.)(.).. (.)(.) The extra piece of evidence is not sufficient to make the prisoner s guilt more likely; it would have to be far more certain.

28 Miscellaneous exercise Bag A contains 1 red ball and 1 black ball, and bag B contains 2 red balls; all four balls are indistinguishable apart from their colour. One ball is chosen at random from A and is transferred to B. One ball is then chosen at random from B and is transferred to A. a. Draw a tree diagram to illustrate the possibilities for the colours of the balls transferred from A to B and then from B to A. b. Find the probability that after both transfers, the black ball is in bag A. a. Tree diagram appears on next slide:

29 b. There are 6 possibilities of which 4 are favourable outcomes, therefore the probability that after both transfers, the black ball will be back in bag A, is given by: 4 6 = 2 3

30 2. The probability that an event A occurs is P A = 0.3. The event B is independent of A and P B = 0.4. a. Calculate P(A or B or both occur). Event C is defined to be the event that neither A nor B occurs. b. Calculate P(C A ), where A is the event that A does not occur. Soln: a. P A or B or both occur = P A + P B P A and B = = 0.58 b. P C = 1 P A or B or both occur = = 0.42 P A = 1 P A = = 0.7 P C A = ( ) ( ) =.. = 0.6

31 3. Two cubical fair dice are thrown, one red and one blue. The scores on their faces are added together. Determine which, if either, is greater: a. The probability that the total score will be 10 or more given that the red dice shows a 6, b. The probability the total score will be 10 or more given that at least one of the dice shows a 6. Soln: a. S(E) = { 1,6, 2,6, 3,6, 4,6, 5,6, 6,6 } F E = { 4,6, 5,6, 6,6 } P score will be 10 or more given that the red dice shows a 6 P score will be 10 or more given that the red dice shows a 6 = () () = =

32 b. S(E) = { 1,6, 2,6, 3,6, 4,6, 5,6, 6,1, 6,2, 6,3, 6,4, 6,5, 6,6 } F(E) = { 4,6, 5,6, 6,4, 6,5, 6,6 } P score will be 10 or more given that at least one of the dice shows a 6 P score will be 10 or more given that at least one of the dice shows a 6 = () () = 4. Half of the A-Level students in a community college study science and 30% study mathematics. Of those who study science, 40% study mathematics. a. What proportion of the A-Level students study both mathematics and science? b. Calculate the proportion of those students who study mathematics, but do not study science. Soln: Let x be the total number of A-Level students. Then, is the proportion of those who study science. is the proportion of those who study math.

33 a. The proportion of the A-Level students study both mathematics and science = =. Hence the proportion is or 20%. b. The proportion of those students who study mathematics, but do not study science = = Hence the proportion is or 10%. Challenge questions: c. What proportion of students study either science or math or both? + + = = or 80% d. What proportion of students study either science or math, but not both? + = = or 40%

34 5. Three friends, Ahmed, Benjamin and Chi live in a town where there are only three cafés. They arrange to meet at a café one evening but do not specify the name of the café. The probabilities that they will each choose a particular café are independent. Ahmed lives close to Café Expresso and so the probability that he will choose to go there is, whereas Café Kola and Café Pepi have equal chances of being visited by him. Benjamin lives a long distance from Café Kola and the probability that he will choose this one is, but he will choose either of the other two cafes with equal probability. Each café has an equal chance of being visited by Chi. a. Show that the probability that the three friends meet at Café Expresso is. b. Calculate the probability that (i) The three friends will meet at the same café. (ii) At most two friends will meet at the same café. Soln: First draw the tree diagram as shown on the following slide:

36 a. From the tree diagram, we see that the probability of the 3 friends meeting at Café Expresso is given by P(Ahmed at Expresso) P(Benjamin at Expresso) P(Chi at Expresso). So, Probability= = = b. (i) The probability that the three friends will meet at the same café is given by P Ahmed at Expresso P Benjamin at Expresso P Chi at Expresso + P Ahmed at Kola P Benjamin at Kola P Chi at Kola + P(Ahmed at Pepi) P(Benjamin at Pepi) P(Chi at Pepi) So, Probability= + + = + + = b. (ii) The probability that at most two friends will meet at the same café is given by 1 P Ahmed at Expresso P Benjamin at Expresso P Chi at Expresso + P Ahmed at Kola P Benjamin at Kola P Chi at Kola + P(Ahmed at Pepi) P(Benjamin at Pepi) P(Chi at Pepi) So, Probability= 1 =

37 6. Two events A and B are such that P A =, P B A = and P B A =. By use of a tree diagram or otherwise, find a. P(A and B) b. P(B) c. P(A B) Soln (without tree diagram): a. P A and B = P A P B A = = b. P B A = P B A P A = = Since P A = 1 P A = 1 = Now, P A + P B P A B + P B A = 1 (See diagram on next slide) P B = 1 + P A B P A P B A P B = 1 + = = = c. P A B = ( ) () = =

39 6. Soln (tree diagram):

40 From the tree diagram: a. P A and B = b. P B = P A B + P A B = + = = = (See diagram) c. P A B = ( ) () = = = =

41 7. Students have to pass a test before they are allowed to work in a laboratory. Students do not retake the test once they have passed it. For a randomly chosen student, the probability of passing the test at first attempt is 1/3. On any subsequent attempt, the probability of failing is half the probability of failing on the previous attempt. By drawing a tree diagram or otherwise, a. Show that the probability of a student passing the test in 3 attempts or fewer is b. Find the conditional probability that a student passed at the first attempt, given that the student passed in 3 attempts or fewer. Soln: a. P student passing the test in 3 attempts or fewer = 1 P(not passing at all) From tree diagram on next slide, P not passing at all = So, P student passing the test in 3 attempts or fewer = 1 = b. Let E be the event that a student passes the test in 3 attempts or fewer. We already know that P E = from part (a). Also, P E P = P E P P = P(P E). P P E = ( ) () = = =

43 8. The probability of event A occurring is P A =. The probability of event B occurring is P B =. The conditional probability of A occurring given that B has occurred is P A B =. a. Determine the following probabilities. (i) P(A and B) (ii) P(B A) (iii) P(A or B or both) (iv) P(A B ) b. Determine P(A occurs or B does not occur) showing your working. Soln: a. (i) P A B = P A B P B = = (ii) P B A = ( ) () = = (iii) P A or B or both = P A + P B P A B = + =

44 (iv) P A B + P A B = 1 P A B = 1 P(A B ) P A B = 1 = 1 = 1 = 1 = b. P A occurs or B does not occur = P(A or B ) P A or B = 1 P B + P A B = 1 + = See diagram belowe:

45 9. a. The probability that an event A occurs is P A = 0.4. B is an event independent of A and P A or B or both = 0.7. Find P(B). b. C and D are two events such that P D C = and P C D =. Given that P C and D occur = p, express in terms of p (i) P(C) (ii) P(D) c. Given also that P C or D or both occur =, find the value of p. Soln: a. P A or B or both = P A + P B P(A B) But, A and B are independent events, therefore, P A B = P A P(B). So, P A or B or both = P A + P B P A P(B) P A or B or both P(A) = P B (1 P A ) () P B = =.. =. =..

46 9. b. C and D are two events such that P D C = and P C D =. Given that P C and D occur = p, express in terms of p (i) P(C) (ii) P(D) Soln: b. (i). P D C = ( ) () (ii). P C D = ( ) () So, P C = ( ) ( ) = So, P D = ( ) ( ) = = 5p = 4p c. P C or D or both occur = P C + P D P(D C) So, = 5p + 4p p = 8p p =

47 10. A batch of 40 tickets for an event at a stadium consists of 10 tickets for the North stand, 14 tickets for the East stand and 16 tickets for the West stand. A ticket is taken from the batch at random and issued to a person, X. Write down the probability that X has a ticket for the North stand. A second ticket is taken from the batch at random and issued to Y. Subsequently, a third ticket is taken from the batch at random and issued to Z. Calculate the probability that (a) Both X and Y have tickets for the North stand (b) X, Y and Z all have tickets for the same stand (c) two of X, Y and Z have tickets for one stand and the other of X, Y and Z has a ticket for a different stand. Soln: P X has a ticket for the North stand = = a. P X and Y have tickets for North stand = =

48 (b) X, Y and Z all have tickets for the same stand: P X, Y and Z have tickets for the same stand = P X, Y and Z have tickets for North stand + P X, Y and Z have tickets for West stand + P X, Y and Z have tickets for East stand P X, Y and Z have tickets for the same stand = + + = =

49 (c) two of X, Y and Z have tickets for one stand and the other of X, Y and Z has a ticket for a different stand: Let E be the event that describes (c). P(X and Y North stand) P(Z has East stand) P E = +P(X and Y North stand) P(Z has West stand) or XY East (Z North/West) or XY West (Z North/East) +P(X and Z East stand) P(Y has North stand) +P(X and Z East stand) P(Y has West stand) or XZ North (Y East/West) or XZ West (Y North/East) +P(Y and Z West stand) P(X has North stand) +P Y and Z West stand P X has East stand 3 or YZ North (X East/West) or YZ East (X North/West) P E = = =

50 11. In a lottery there were 24 prizes allocated at random to 24 prize winners. Ann, Ben and Cal are three of the prize winners. Of the prizes, 4 are cars, 8 are bicycles and 12 are watches. Show that the probability that Ann gets a car and Ben gets a bicycle or a watch is. Giving each answer either as a fraction or a decimal correct to 3 significant figures, find a. the probability that both Ann and Ben get cars, given that Cal gets a car b. the probability that either Ann or Cal (or both) gets a car c. the probability that Ann gets a car and Ben gets a car or a bicycle d. the probability that Ann gets a car given that Ben gets either a car or a bicycle Soln: a. + = b. P neither Ann nor Ben gets car = = P Either Ann or Ben or both get cars = 1 =

51 c. + = = d = = Remaining solutions to be uploaded soon. Keep checking back. Source: John Gabriel

Topic 5: Probability. 5.4 Combined Events and Conditional Probability Paper 1

Topic 5: Probability. 5.4 Combined Events and Conditional Probability Paper 1 Topic 5: Probability Standard Level 5.4 Combined Events and Conditional Probability Paper 1 1. In a group of 16 students, 12 take art and 8 take music. One student takes neither art nor music. The Venn