MATH 382 Conditional Probability Dr. Neal, WKU We now shall consider probabilities of events that are restricted within a subset that is smaller than the entire sample space Ω. For example, let Ω be the space of all students, with F being female students, and consider the event G of having a GPA of at least 3.0. Then P(G) represents the probability that a student (from all of Ω ) has a GPA of at least 3.0. But suppose we want to consider just female students. Then we use P(G F) to denote the probability that a female student has a GPA of at least 3.0. This conditional probability can be stated in several ways, such as (i) Given that a student is female, what is the probability of having a GPA of at least 3.0? (ii) What is the probability, among just females, of having a GPA of at least 3.0? We formally define this concept in the following way: Definition. Let D be an event with 0 and let C be another event. The conditional probability of C given D is defined by P(C D) P(C D) / P( D). (The second (given) set is the reduced sample space.) Example 1. Among all students, 60% are female (event F ), 70% have a GPA of at least 3.0 (event G ), and 48% are females with a GPA of at least 3.0. Let M F c be the males. Compute and explain P(G F), P(F G), P(M G), P(G c M), and P(M G c ). Solution. We first show all information on a Block diagram: G G c G G c F 48% 60% F 48% 12% 60% M M 22% 18% 40% 70% 100% 70% 30% 100% P(G F) P(G F) P(F ) 48 0.80 80% of females have a GPA of at least 3.0. 60 P( F G) P(F G) 48 P( M G) 0.6857 and P(M G) 22 0.3143 Among P(G) 70 P(G) 70 just those with a GPA of at least 3.0, 68.57% are female and 31.43% are male. P(G c M) P(Gc M) P(M) P(M G c ) P( M Gc ) P(G c ) 18 0.45 45% of males have a GPA under 3.0. 40 18 30 0. 60 Among those with a GPA under 3.0, 60% are male.
Example 2. Flip a fair coin three times. What is the probability of (a) All Heads given that there is at least one Head? (b) Exactly 2 heads given that the first is a Head? Solution. Let X count the number of Heads. Then consider the sample space of all 8 outcomes, and just look at the reduced sample space. HHH HHT HTH THH HTT THT TTH TTT (a) Given that there is at least one Head, then we are reduced to 7 possibilities. Out of these 7, only 1 has all heads; thus, P(X 3 X 1) 1. Or, because {X 3} {X 1}, 7 we also could compute the value as P(X 3 X 1) P( X 3 X 1) P( X 1) P( X 3 ) P( X 1) 1 / 8 7 / 8 1 7. (b) Let X 1 be the outcome of the first roll. Given that X 1 H, we are reduced to 4 possibilities: HHH HHT HTH HTT. Of these 4 outcomes, 2 of them have exactly 2 Heads. Thus, P(X 2 X 1 H) 2 4. Alternately, {X 2 X 1 H} {HHT, HTH}, so P(X 2 X 1 H) P( X 2 X 1 H) P( X 1 H) 2 / 8 1 / 2 2 4. Multiplication Rule Given events A, B, with P(A) 0 and P(B) 0,we have P(B A) P( A B) P( A) and P(A B) P( A B). P( B) So we can write P(A B) as P(A B) P( A) P(B A) or P(A B) P( B) P( A B) In other words, what is the chance of both A and B happening? First we need A, then we need B given that A has happened: P(A) P(B A). (Or we need B and then we need A given that B has happened: P(B) P(A B). ) Note: When A and B are known to be independent, then P(A B) P( A) P(B). But iwhen there is dependence, then A affects B so that P(B) P(B A). So when we have dependent events, we need P(A B) P( A) P(B A).
P(A 1 A 2 A 3... A n ) More generally, we can write P( A 1 ) P( A 2 A 1 ) P( A 3 A 1 A 2 )... P(A n A 1 A 2... A n 1 ). Example 3. If drawing 5 cards in sequence, what is the probability of drawing a Heart, then a Spade, then a Spade, then a Club, then a Heart if drawing (a) with replacement and reshuffling? (b) without replacement? Solution. (a) With replacement and reshuffling, then each draw is independent and the probability of each suit on each draw is 1/4. Thus, P(H S S C H) P(H) P(S) P(S) P(C) P( H) (1 / 4) 5. (b) Without replacing cards, we have dependent draws, so we now must use the general multiplication rule: To start, there are 52 cards and 13 in each suit. But these numbers reduce by one as each card/suit is drawn. Thus, P(H S S C H) P(H) P(S H) P(S H S) P(C H S S) P( H H S S C) 13 52 13 51 12 50 13 49 12 0. 001 48 Sometimes information is given in terms of conditional probabilities: Example 4. Among WKU students, 40% smoke (event S ) and 65% drink (event D). But among smokers, 80% drink. Find the probabilities of (a) Smoking among just drinkers (b) Not smoking given that one doesn't drink (c) Doing both given that one does one or the other (d) Not doing either given that one doesn't do both. Solution. The information given is P(S) 0. 40, 0.65, and P(D S) 0.80. To complete a Block diagram, we need P(S D) P(S) P(D S) 0.40 0.80 0.32 which is really just taking 80% of the 40% of students who smoke: D D c D D c S 0.4 0.8 0.40 S 0.32 0.08 0.40 S c S c 0.33 0.27 0.60 0.65 1 0.65 0.35 1
Now we can compute the desired conditional probabilities: (a) P(S D) P(S D) P( D) 32 65 0.4923. (b) P(Sc D c ) P( Sc D c ) P(D c 27 ) 35 0.7714 (c) Because S D S D and P(S D) 0. 73, we have P(S D S D) P(S D) P(S D) 32 73 0.438. (d) Now S c D c (S D) c ; thus, P(S c D c (S D) c ) P(Sc D c ) P((S D) c ) 27 68 0.397. Special Cases (i) C and D are disjoint (ii) C is contained in D C D C D Ω Ω P(C D) P(C D) P( ) 0 P(C D) P(C D) P(C) Given D, there is then no chance of C. (iii) D is contained in C Ω P(C D) C D P(C D) 1 Given D, then C happens also. P(C D) P(D C) (iv) C and D are independent P(C D) P( D C) P(C) and P(C) P( D) P(C) P(C) P(C) C and D have no affect on each other.
Example 5. Among the population of senior citizens, 90% need glasses. Among those needing glasses, 75% need bifocals. (a) What percentage of senior citizens need bifocals? (b) If one doesn't need bifocals, what is the probability of not needing glasses? Solution. Let P(G) 0.90 be the probabilty of needing glasses and P(B G) 0.75 be the probabilty of needing bifocals given that one needs glasses. Because B G, we have P(B G) 0.75 P(B G) P(B) P(G) P(G) P(B) 0.90. Hence, P(B G) P( B) 0. 90 0.75 0.675. B B c G 0.675 0.225 0.90 G c 0 0.10 0.10 0.675 0.325 1 Because B G, then G c B c. Thus, P(G c B c ) P(Gc B c ) P(B c ) P(Gc ) P(B c ) 0.10 0. 325 4 13 0.3077. Example 6. Deal 5 cards. What is the probability of having at least 3 Hearts given that you have at least 2 hearts? Solution. Let X denote the number of Hearts. Then {X 3} {X 2}; hence, P(X 3 X 2) P(X 3 X 2) P(X 2) P(X 3) P( X 3) + P( X 4) + P( X 5) P( X 2) 1 P( X 0) P( X 1) C(13, 3)C(39,2) + C(13, 4)C(39,1) + C(13, 5)C(39, 0) C(52, 5) C(52, 5) C(13, 0)C(39,5) + C(13, 1)C(39, 4) C(52, 5) 241, 098 953, 940 0.25274.
Example 7. Suppose 60% of students are female. Among the females, the probability of having an automobile is 0.65. Among the males, the probability of not having an automobile is 0.25. Among all, 40% of the autos are trucks. If one has a truck, there is an 80% chance of being male. (a) Among all students, what is the probability of having an automobile? (b) If one has an auto, then what is the probability of being male? (c) If one does not have an auto, then what is the probability of being female? (d) What percentage of the students have a truck? (e) If one does not have a truck, what is the probability that one does not have an automobile? (f) Overall, what percentage are male and have a truck? (g) Is having a truck independent of sex? Explain. Solution. We first use the information given to complete a block diagram: Let F be the females, M be the males, A be those having an auto, and T be those having a truck. Then P(F A) P(F ) P( A F ) 0.60 0.65 0.39. P(M A c ) P(M) P( A c M) 0.40 0.25 0.10 A A c F 0.60 0.65 0.39 0.21 0.60 M 0.30 0.40 0.25 0.10 0.40 0.69 0.31 1.00 (a) From the Block diagram, we obtain P(A) 0.69, which also comes from P(A) P(F) P(A F) + P(M) P(A M) 0.60 0.65 + 0.40 0.75 0.69. (b) P(M A) P( M A) P( A) 0. 30 0. 69 10 23 0.43478. (c) P(F A c ) P(F Ac ) P( A c ) 0.21 0.31 21 31 0.6774 (d) Because 40% of the autos are trucks, we have P(T) 0.40 0.69 0.276, so 27.6% have a truck. This result also follows from 0.40 P(T A) P(T ) P( A) P(T) 0. 69. (e) Because T A, then A c T c. Thus, P(A c T c ) P( Ac ) P(T c ) 0.31 0. 724 0.428.
(f) Overall, what percentage are male and have a truck? P(M T ) P(T) P( M T) 0. 276 0.80 0.2208, so 22.08% of the students are males with a truck. (g) Is having a truck independent of sex? No; we have P(M T ) 0. 2208, but P(M) P( T ) 0. 40 0.276 0.1104, so having a truck is dependent on the sex of the student. P(T M) Also note that P(T) 0.276 while P(T M) 0.2208 0.552. Thus, P(M) 0.40 males are more likely to have a truck than a generic student in the whole population. As shown in Part (g), conditional probabilities can be used to check for independence. The formal definition of A and B being independent is that P(A B) P( A) P(B). But intuitively, it should be the case that A and B are independent provided they have no affect on each other; that is, it should be the case that P(B A) P(B) and P(A B) P(A). We prove this next: Proposition 5. Suppose events A and B are independent with P(A) 0 and P(B) 0. Then: (a) P(A B) P(A) and P(B A) P(B). (b) If P(A B) P( A) or P(B A) P(B), then A and B must be dependent. Proof. (a) By independence, we know P(A B) P( A) P(B). Thus P(A B) P( A B) P( B) P(A) P(B) P(B) P( A), and P(B A) P(B A) P( A) P(B) P(A) P(A) P(B). Part (b) is simply the contrapositive of (a) whicj is logically equivalent. Note: We always have P(A) + P( A c ) 1. It is also true that P(A B) + P( A c B) 1; however, in general, P(A B) + P( A B c ) 1.
Exercises 1. Among adults in Kentucky, 40% smoke, 55% drink, and 59% smoke or drink. (a) Show all information on a block diagram. (b) Compute the probability of (i) Not smoking given that one doesn't drink (ii) Doing exactly one of smoking or drinking given that one either smokes or drinks. (c) Among drinkers, what percentage smoke? (d) Among non-smokers, what percentage don't drink? (e) Determine if drinking is independent of smoking among adults in KY. (f) Compared to the general population of adults in KY, determine if a smoker is more likely, less likely, or equally likely to be a drinker. (g) Compared to the general population of adults in KY, determine if a non-drinker is more likely, less likely, or equally likely to be a smoker. 2. A class of 20 students has 12 Females and 8 males. Six students are chosen at random all at once. (a) What is the probability that there are at least 4 females chosen? (b) Given that there are at least 4 females chosen, what is the probability that all are female? 3. In a large apartment complex, tenants are allowed to have one pet per apartment. Among all apartments, 75% have a pet. If one has a pet, there is a 60% chance that it is a dog. (a) What percentage of these apartments have a dog? (b) What is the probability of not having a pet given that one doesn't have a dog? (c) Show all information on a block diagram. 4. Let A and B be events with P(B) 0. Prove that P(A B) + P( A c B) 1. 5. Let A and B be events with 0 < P(A) < 1: (a) Suppose P(B A) P(B). Prove that A and B are independent. (b) Prove that A and B are independent if and only if P(B A) P(B A c ).
6. Draw two cards at random from a shuffled deck, in order, without replacement. Compute the probability of drawing (a) Two spades (b) A Spade, then a Club (c) A spade, then a Queen (d) No Spades (e) Exactly one Spade. (Hint: Consider two cases.) 7. Consider the numbers 1, 2, 3, 4, 5, 6 placed in a hat. You and an opponent draw a single number at random one at a time without replacement with you drawing first. The winner is the person who draws the 6. (a) What is the probability that you win? (b) If instead you roll a single die until one of you rolls the 6, then what is the probability that you win if you roll first?