Probability Rules: E F S E - Texas A&M University

1 rules for ProbabilityProbability Rules: For any events,EandF, in a sample spaceS, we have(1) 0 P(E) 1(2)P(S) = 1(3)P( ) = 0(4)P(E F) =P(E) +P(F) P(E F) (Union Rule for Probability ) (E F) =n(E) +n(F) n(E F), divide both sides byn(S), we haven(E F)n(S)=n(E)n(S)+n(F)n(S) n(E F)n(S).We complete the proof by definition of Probability .(5) If E and F are mutually exclusive events, thenP(E F) =P(E)+P(F). (by part (4),n(E F) = 0)(6)P(EC) = 1 P(E) andP(E) = 1 P(EC) (Complement (S) = mutuallyexclusive)Example two events of an experiment with sample sample spaceS. SupposeP(E) = ,P(F) = (E F) = Compute the following:(a)P(E F); (b)P(FC); (c)P(EC FC); (d)P(EC F).Solution.(a)P(E F) =P(E) +P(F) P(E F) = (b)P(FC) = 1 P(F) = (c)P(EC FC) =P((E F)C) = 1 P(E F) = (d)P(EC F) =P(F) P(E F) = (E) = (F) = exclusive, what isP(EC FC)? (EC FC) =P((E F)C) = 1 P(E F) = 1 (P(E) +P(F)) = you are given the following Probability distribution for a sample spaceS={s1, s2, s3, s4, }.

2 Outcomes1s2s3s4s5s6 Probability310120112110 SupposeE={s1, s4, s5},F={s2, s3},G={s2, s5}, andP(E) =1120. Fill in the missing Probability inthe table and then calculate the )P(F G)andP(E F)b)P(GC)andP(E G)1c)P(FC GC) (s4) =P(E) P(s1) P(s5) =1120 310 112=16andP(s4) = 1 P(s1) P(s2) P(s3) P(s5) P(s6) =310.(a)P(F G) =P(s2) = (E F) =P( ) = 0.(b)P(GC) = 1 P(G) = 1 310 112= (E G) =P({s3, s6}C) = 1 P(s3) P(s6) = 1 120 110=1720.(c)P(FC GC) =P((F G)C) =P({s2}C) = 1 P(s2) = experiment consists of flipping a fair coin and rolling a fair six-sided ) Find the Probability of flipping a heads and rolling an even ) Find the Probability of flipping a heads or rolling an even ) Find the Probability of rolling an even number or rolling ) Find the Probability of not rolling a ) Find the Probability that neither a head is flipped nor an even number is sample spaceS={(H,1),(H,2),(H,3),(H,4),(H,5),(H ,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)} .

3 (a) The event{flipping a heads and rolling an even number}={(H,2),(H,4),(H,6)}.So the Probability is312=14.(b) The eventE={flipping a heads}={(H,1),(H,2),(H,3),(H,4),(H,5),(H ,6)}.The eventF={rolling an even number}={(H,2),(H,4),(H,6),(T,2),(T,4),( T,6)}.SoE F={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),( T,2),(T,4),(T,6)}andP(E F) =912=34.(c) The eventG={rolling 5}={(H,5),(T,5)}.SoG F={(H,2),(H,4),(H,6),(T,2),(T,4),(T,6),( H,5),(T,5)}andP(E F) =812=23.(d)P(GC) = 1 P(G) =56.(e)P(EC FC) =P((E F)C) = 1 P(E F) = table below gives the number of students of each classification who are majoring andnot in business in a class of 110 student is randomly selected from this class. Find the Probability of the following ) The student is not a ) The student is a lower classman (freshman or sophmore).c) The student is an upper classman non-business ) The student is a business major or a (a) Let the eventA={The students is a junior}.

4 SoP(A) =n(A)n(S)=35110= (AC) =1 P(A) =1522.(b) Let the eventB={The student is a lower classman}. Son(A) = 18 + 20 = 38. Then,P(B) =38110=1955.(c) Let the eventC={The student is a upper classmain and non-business major}. Then,n(C) =15 + 25 = 40. Then,P(C) =40110=411.(c) Let the eventsD={The student is a business major}andE={The student is a sophmore}. Son(D) = 59,n(E) = 20,n(D E) = 17. Thenn(D E) =n(D) +n(E) n(D E) = 62. Thus,P(D E) =62110= :The odds of an eventEare defined to be the ratio ofP(E) toP(EC), orP(E)P(EC).Example the Probability that it will rain tomorrow is , what are the odds that it will raintomorrow? What are the odds that is will not rain tomorrow? the eventE={It rains tomorrow}. So the odds that it will rain areP(E)1 P(E)= odds that it will not rain areP(EC)P(E)= Probability from Odds:Suppose that the odds for an eventEoccurring is given asabora:b, thenP(E) =aa+ odds of Whipper Snapper (a horse) winning at hte Kentucky Derby are 5 to 3.

5 Whatis the Probability that Whipper Snapper will win? What is the Probability that Whipper Snapper will notwin? the eventE={Whipper Snapper will win}. SoP(E)1 P(E)= ,P(E) =55 + 3= (EC) = 1 P(E) = : Conditional ProbabilityExample a six-sided dice and observe the number of dots on the top ) Let the eventE={the number 4 or 3}. SoP(E) =26= ) Assume that we know the number we observe is always even (Maybe using some high-techies). Whatis the Probability we observe the number 3 or 4? there are only three outcomesF={the number 2 or 4 or 6}. We could only observe4. So the reasonable Probability should be one out of three, 1/3. Consider the general probabilityframework we discussed before. The sample space has been changed! The new sample space isFand the desired event should beE F. Therefore, our Probability under the conditionFisP(E|F) =n(E F)n(F)= Probability :LetEandFbe two events in a sample spaceS.

6 The Probability thatEoccurs given thatFhas occurred is defined to beP(E|F) =n(E F)n(F)=n(E F)/n(S)n(F)/n(S)=P(E F)P(F).Example pair of fair six-sided dice is rolled. One is red and the other is ) What is the Probability that the sum is equal to 5 given that a 2 is rolled?b) What is the Probability that a 2 is rolled given that the sum is less than 5? thatn(S) = 36. Let the eventE={the sum is equal to 5}={(1,4),(2,3),(3,2),(4,1)}and the eventF={2 is rolled}={(2,1),(2,2),(2,3),(2,4),(2,5),( 2,6),(1,2),(3,2),(4,2),(5,2),(6,2)}.SoE F={(2,3),(3,2)}and(a)P(E|F) =n(E F)n(F)= (b)P(F|E) =n(E F)n(E)=24= (E) = ,P(F) = , andP(E F) = Finda)P(E|F); b)P(FC|E); c)P(F|EC); d)P(E F|F); e)P(EC F|E). )P(E|F) =P(E F)P(F)= )P(FC|E) =P(FC E)P(E)=P(E) P(E F)P(E)= )P(F|EC) =F ECP(EC)=P(F) P(F E)1 P(E)= )P(E F|F) =P((E F) F)P(F)=P(F)P(F)= )P(EC F|E) =P((EC F) E)P(E)=P( )P(E)= Rule:IfEandFare two events in a sample spaceSwithP(E)>0 andP(F)>0,thenP(E F) =P(F)P(E|F) =P(E)P(F|E).

7 Example 11(ex. 38).Two machines turn out all the products in a factory, with the first machineproducing 40% of the product and the second 60%. The first machine produces defective products 2% ofthe time and the second machine 4% of the ) What is the Probability that a defective product is produced at this factory and it was made on thefirst machine?b) What is the Probability that a defective product is produced at this factory?c) Given a defective product, what is the Probability it was produced on the first machine? {Products from the first machine}andE2={Products from the second machine}.We also letF1={Defective products from the first machine}andF2={Defective products from the second machine}.SoP(E1) = ,P(E2) = ,P(F1|E1) = andP(F2|E2) = ) LetE={Defective products at this factory}. ThenP(E E1) =P(E|E1)P(E1) = = )P(E) =P(F1) +P(F2) =P(F1 E1) +P(F2 E2) =P(E1)P(F1|E1) +P(E2)P(F2|E2) = )P(E1|E) =P(E1 E)P(E)=P(F1)P(E)=P(F1|E1)P(E1)P(E)= Events:We say that two eventsEandFare independent if the outcome of the onedoes not affect the outcome of the other.

8 So the conditional probabilityP(E|F) does not depend onP(F) (The event E happens no matter whether the event F has happened), that isP(E|F) =P(E).SimilarlyP(F|E) =P(F).Note that ifP(E)>0 andP(F)>0,P(E) =P(E|F) =P(E F)P(F),andP(F) =P(F|E) =P(E F)P(E).SoP(E F) =P(E)P(F). This is independent events exclusive vs if the givens eventsEandFare )P(E) = ,P(F) = , andP(E F) = )P(E FC) = ,P(E F) = , andP(EC F) = (a)P(E)P(F) = (E F). They are not independent.(b)P(E) =P(E FC) +P(E F) = (F) =P(F EC) +P(E F) = SoP(E)P(F) = =P(E F). They are the eventsEandFare independent withP(E) = (F) = FindP(E F). (E F) =P(E) +P(F) P(E F) =P(E) +P(F) P(E)P(F) = + = fair coin is flipped three times. LetEbe the event at most one head andFtheevent at least one head and at least one tail. Are there two events independent? {At least one head}andF={At least one head and at least one tail}.

9 ThenP(E) =48=12andP(F) =68= (E F) =38. SoP(E)P(F) =38=P(E F). Thus, theyare set of events{E1, E2, , En}is said to be independent if, for anykof these events, the probabilityof the intersection of thesekevents is the product of the probabilities of each of thekevents. Thusmust hold for anyk= 2,3, , example, if we want to verify the three eventsE,FandGare independent, we need to verifyP(E F) =P(E)P(F),P(F G) =P(F)P(G),P(G E) =P(G)P(E),P(E F G) =P(E)P(F)P(G).Example ,FandGare independent,a) Show thatECandFare (EC)P(F) =P(F) P(E)P(F) =P(F) P(E F) =P(EC F).b) Show thatEC,FandGare definition,P(EC F) =P(EC)P(F),by part (a)P(F G) =P(F)P(G),P(G EC) =P(G)P(EC),similar argument by part (a)P(EC F G) =P(EC)P(F)P(G) = (1 P(E))P(F)P(G)=P(F)P(G) P(E)P(F)P(G)=P(F G) P(E F G) =P(EC F G).7