Probability and Counting


 Kelley Burns
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1 Probability and Counting Basic Counting Principles Permutations and Combinations Sample Spaces, Events, Probability Union, Intersection, Complements; Odds Conditional Probability, Independence Bayes Formula Random Variable, Distribution, Expectation Basic Counting Principles Sets, Operations on Sets Addition Principle Venn Diagrams Multiplication Principle
2 Some Terminology A SET is a COLLECTION of objects called ELEMENTS A={a,b,m,n,1,2,3} 1 A 1 is an element of A C={1,2,3} C A C is a subset of A D={ { },{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3} } E={ } EMPTY SET POWER SET UNION: A B all elements which are in A or B (or both) but no repeats! { 1,2,3 } { 2,a,4} = { 1,2,3,4, a} INTERSECTION: A B all elements that are in both, A and B {1,2,3} {2, a,4} = {2}
3 Complements: Universal Set = Set of all things to be considered for this case U = all students in class A = students getting an A B = students getting a B A = complement of A, all students not getting an A A = U \ A U minus A Addition Principle Counting elements in a set: n(a)= number of elements in A (distinct) n({1,z,2,3})= 4 F = set of female students in this room M= set of male students in this room M F set of students in this room n(m F) = n(m) + n(f)
4 A = B = set of business majors registered for class set of students present today A B = set of business majors or students present today A 1 B = set of business majors present today n(a B) = n(a) + n(b)  n(a B) Addition Principle for Counting For any two sets A and B n(a B) = n(a) + n(b) n(a B) If A and B are disjoint, i.e. A B = or { } n(a B) = n(a) + n(b)
5 Practical Problem U students in this class 40 A students owning car B students owning computer 30 students own a car 20 students own a computer 15 own both How many have neither? Venn Diagram U A B
6 Assigning Students Numbers Suppose each student is assigned a 5 digit number. How many different numbers can be created? Each digit is 0,1,2,3,4,5,6,7,8,9 Ten possibilities for each 10*10*10*10*10=10 5 =100,000
7 More interestingly A) Suppose not digit is repeated First slot: 10 Possibilities Second 9 Possibilities Third 8 Possibilities Fourth 7 Possibilities Fifth 6 Possibilities TOTAL = 10*9*8*7*6 = B) No two adjacent digits are equal First slot: 10 Possibilities 2,3,4,5: 9 Possibilities TOTAL = 10*9*9*9*9 = Multiplication Rule 1. If two operations O 1 and O 2 are performed in order, with N 1 and N 2 possible outcomes, respectively, then there are N 1 *N 2 possible combined outcomes. 2. For O 1,O 2,,O k operations with N 1,N 2,,N k possible outcomes, then there are N 1 *N 2 * *N k possible combined outcomes
8 Permutations and Combinations FACTORIAL PERMUTATIONS COMBINATIONS APPLICATIONS Factorial Counting without replacement involved 25*24*23* *2*1 and similar products, multiplying several numbers, each one less than the previous.
9 Beginning Lotto: Pick 6 numbered balls out of 49 (numbered consecutively 1,2,3,,49) without replacing them. How many ways can this be done (the order is important here) First ball: 49 choices Second : 48 Third: 47 Fourth: 46 Fifth: 45 Sixth: 44 Total: 49*48*47*46*45*44 = 10,068,347,520 So there are over 10 billion ways to pick the six balls. But the order is not important when playing Lotto etc. are the same and should not be counted separate. We will get back to that soon.
10 Factorial For a natural number n, n!=n*(n1)*(n2)* *2*1 0!=1 n!=n*(n1)! Read n factorial. Examples: 6! = = 720 7!=7 (71)!=7 (6!)=7 720= ! = 43! (43!) = = ! 49! = = = 13,983,816 43!6! 6! 720
11 Permutations Recall the lotto example: Order was not important! Question: How many was can we rearrange or permute the numbers We can choose any one of the 6 for the first slot, any of the remaining 5 for the second etc. 1st 6 2nd 5 3rd 4 4th 3 5th 2 6th 1 Total 6!=720.
12 There is nothing particular about the numbers , any six distinct objects can be arranged in 6! different ways. A Permutation of a Set of Objects A permutation of a set of distinct objects is an arrangement of the objects in a specific order without repetition. Number of Permutations of n Objects The number of permutations of n distinct objects without repetition, denoted P n,n, is given by P n,n = n*(n1)*(n2)* *2*1=n!
13 A permutation of n Objects r at a Time In our Lotto example we have 49 numbers available but pick only 6 at a time: 49 choices for the first 48 for the second 44 for the sixth 49! 49! = = = 43! (496)! P 49,6 A Permutation of n Objects, r at a time A permutation of a set of n distinct objects taken r at a time without repetition is an arrangement of the r objects in a specific order
14 The number of Permutations of n Objects taken r at a time The number of permutations of n distinct objects taken r at a time without repetition is given by P P n,r n,r = n (n  1) (n r + 1) = n! (n  r)! 0 r n P n,n r = n! factors Combinations In our Lotto example the order of the six numbers did not matter. So we can NOT use the permutation formula. Combination of n Objects taken r at a time A combination of a set of n distinct objects taken r at a time without repetition is an r element subset of a set of n objects. The arrangement of the elements does NOT matter.
15 Simple Example: How many ways can 2 out of 3 paintings of an artist be selected for shipment to an exhibition? Let the paintings correspond to {A,B,C} First choice: A B C Second: B or C A or C A or B Ordered: (A,B) ; (A,C) (B,A) ; (B,C) (C,A) ; (C,B) There are 6 permutations, but only 3 combinations: {A,B};{A,C};{B,C} Number of Combinations of n objects taken r at a time The number of combinations of n distinct objects taken r at a time without repetition is given by n Pn,r n! Cn, r = = = 0 r n r r! r!(n  r)!
16 How many 5 card hands can be drawn from a 52 card deck? 52 52! 52! C 52,5 = = = 5 5!(525)! 120 (47!) = 2,598, = Permutation: ORDER MATTERS Combination: ORDER DOES NOT MATTER Sample Spaces, Events and Probability EXPERIMENTS SAMPLE SPACES AND EVENTS PROBABILITY OF AN EVENT EQUALLY LIKELY ASSUMPTION
17 Experiments We need: Mathematical Model for Probability Random Experiments: Do not yield same result if repeated Result of Experiment = Outcome Sample Spaces and Events Experiment = rolling a six sided die Outcomes: one dot up two dots up etc. Possible outcomes of one experiment: Simple Outcomes Simple Events S=,,,,,
18 Sample Spaces and Events S is a sample space for an experiment if in each trial one and only one of the elements of S can occur as outcome. The elements of S are called simple events or simple outcomes. An event E is any subset of S, including the empty set and S itself. E is a simple event if it contains exactly one element of S, it is a compound event if it contains more than one element. We say the event E occurs if any of the simple events in E occur. For dice we could think of the die having the numbers {1,2,3,4,5,6} written on each side (instead of dots), then S= {1,2,3,4,5,6}. A simple event would be E={3} (rolling a 3) or E={6} (rolling a 6) A compound event E={1,3,5} (rolling and odd number) this happens if we roll a 1 or a 3 or a 5
19 Coin Tossing: Tossing one coin: Once: Head or Tail S={H,T} Twice: S={HH,HT,TH,TT} Tossing two coins(nickel and dime): Once: Start Nickel Dime Combined H T H T H T HH HT TH TT The sample space also depends on what question in which we are interested: Number of heads when tossing two coins: S={0,1,2} Do both coins show the same side: S={yes,no}
20 Choosing the Sample Space: There is NO ONE correct sample space for a given experiment. We need to include enough detail to answer all questions of interest regarding the outcomes of the experiment. When in doubt choose a sample space with more rather than less elements. Rolling a die and tossing a coin: H (H,1) (H,2) (H,3) (H,4) (H,5) (H,6) T (T,1) (T,2) (T,3) (T,4) (T,5) (T,6) S={ (H,1),(H,2),(H,3),(H,4),(H,5),(H,6), (T,1),(T,2), (T,3),(T,4),(T,5), (T,6) }
21 Probability of an Event We need a way to compare different experiments such as two lotteries, say Lotto Kentucky and Power Ball. Which one (A) gives us a better chance of winning (B) is more likely to produce more money To answer (A), we need a NUMBER that is independent on the exact nature of the experiment and measures the likelihood of winning. Number of good outcomes versus Number of all possible outcomes Probability of rolling a 3 with one fair six sided die is 1 in 6 or 1/6.
22 Probabilities for Simple Events Given a sample space S={e 1,e 2,,e n } with n simple events e i we assign a real number, denoted P(e i ), called the probability of e i to each simple event. These numbers have to meet only the following conditions: 1) 0 < P(e i ) < 1 for i=1,2,,n 2) P(e 1 )+P(e 2 )+ +P(e n )=1 Equally Likely Assumption If the experiment is such that each simple event is equally likely (fair coin, fair die, fair poker dealer, but not horse racing), then Probability of a simple event with equally likely assumption: Sample space S={e 1,e 2,,e n }, with n elements, then under equally likely assumption we have P(e ) = 1 i n for i=1,2,,n
23 Rolling a die: is it fair or not? Experiment: Roll it 600 times If fair: About 100 times each of 1,2,3,4,5,6 Empirical Probability Approximation P(e ) i Frequency of occurrence of e Total number of trials i = f(e ) i n (The larger n is the better the approximation) Probability of an arbitrary event E Given probabilities for simple events, we define the probability of an event E as (A) (B) (C) (D) E is empty set, then P(E)=0 E is simple, then P(E) is known E is compound, then P(E) is the sum of all P(e i ) for e i in E (over all simple events in E) E=S (the sample space) then P(E)=1 (Note: this is a consequence of C)
24 theorem 1 Probability of E under equally likely If we assume each simple event in the sample space S to be equally likely, then P(E) = Number of elements in E Number of elements in S = n(e) n(s) Page 403 #20 Suppose 6 people check their coats in a checkroom. If all claim checks are lost and the 6 coats are randomly returned, what is the probability that all people will get their own coats back?
25 Solution The experiment is drawing 6 coats without replacement from 6 available coats. Person A gets the first coat, Person B the second etc. So order does matter and we need permutations of 6 coats 6 at a time, P 6,6 For the probability we need to know how many correct assignments there are. Only one! P(E) = 1 P 6,6 = 1 6! = Union, Intersection, Complement of Events Odds UNION and INTERSECTION COMPLEMENT of an EVENT ODDS APPLICATIONS to EMPIRICAL PROBABILITY
26 Union and Intersection A, B events in (i.e. subsets of) sample space S, then A B ( A union B ) and A B ( A intersect B ) are A B = {e S e A or e B} A B S A B = {e S e A and e B} S The event A or B is A B The event A and B is A B Example: Rolling a die, S={1,2,3,4,5,6} a) E=Rolling an odd number or a number greater than 3. A={1,3,5} B={4,5,6} E=A B={1,3,4,5,6} P(E)=n(E)/n(S)=5/6 b) F=Rolling an odd number and a number greater than 3. A={1,3,5} B={4,5,6} F=A B={5} P(F)=n(F)/n(S)=1/6
27 Recall: n(a B) = n(a) + n(b)  n(a And since n(a B) n(a) + n(b)  n(a B) P(A B) = = n(s) n(s) Probability of a Union of two Events: P(A B)=P(A)+P(B)P(A B) B) If A and B are mutually exclusive (A B)= ) P(A B)=P(A)+P(B) Complement of an Event S={e 1,e 2,,e n } E and E subsets of S such that E E =S E E = then E is the complement of E in S E = set of all elements of S NOT in E.
28 E S E Recall that P(S)=1 and P(E E )=P(E)+P(E ) Complements P(E)=1P(E ) P(E )=1P(E)
29 Birthday Problem Suppose everyone has one of 365 birthdays (no leap years). We have 39 students registered what is the probability that at least two have the same birthday (only day not year)? S=possible birthdays for 39 people: n(s)=365*365* *365 = E=at least two people have same birthday E =everyone has a different birthday To determine n(e ), we note that the first person can have any of 365 birthdays, but the second only 364 (not the same as the previous person) and the third only 363 (not same as all previous persons) etc. n(e )=365*364*363* (36538) (39 terms) =365!/(36539)! with the equally likely assumption 365! P(E )=n(e )/n(s)= 39 = (36539)! P(E)=1P(E )=0.878
30 From Probability to Odds Odds for E and odds against E If P(E) is the probability of event E, then Odd for E= P(E) 1P(E) = P(E) P(E') P(E) 1 Odds against E= P(E') P(E) P(E) 0 Probability measures: good events divided by all possible events Odds measure: good events divided by bad events or Probability of winning divided by Probability of loosing
31 From Odds to Probability If the odds for the event E are a/b, then the probability of E is P(E)= a a+b Fair Game If the odds for the event E are a/b, then we say the game is fair if your bet of $a is lost if E does not happen, but you win $b if E does happen.
32 Single Roll of Two Fair Dice What is the probability (what are the odds) of rolling a sum of 8? 1st d i e r o l l roll of the second die There are 5 ways of getting an 8, but there are 36 total possibilities, thus P(E)= 5 36 = The odds for E are 5 5 P(E) 1P(E) = 36 = 36 =
33 If you bet $5 that a sum of 8 turns up, how much should the house pay to make this a fair game? Since the odds are 5:31 the house should pay $31 if a sum of 8 comes up. Applications to empirical probability Odds for E is number of elements in E divided by number of elements in S but NOT in E 1000 people are surveyed, 500 tried brand A 600 tried brand B and 200 tried both. What are the empirical odds for the event E that one person has tried at least one of the brands?
34 n(a B) = n(a) + n(b)  n(a B) = = 900 number of people who haven t tried either = 100 Therefore the odds for E are Number of people who tried either number of people who tried neither 900 = 9 : "nine to one" empirical probability is n(a B) = n(s) = = 0. 9 Conditional Probability, Intersection, Independence CONDITIONAL PROBABILITY INTERSECTION OF EVENTS (PRODUCT RULE) PROBABILITY TREES INDEPENDENT EVENTS SUMMARY
35 Probability of someone carrying a bomb on an airplane = p 1 Probability of two people carrying a bomb on an airplane = p 2 p 2 << p 1 Should you always carry a bomb? NO! Conditional Probability The probability of a second person carrying a bomb, given that you already brought one, is the same as the probability any one person carrying on a bomb. P(second bomb first bomb)=p(one bomb) probability of a second bomb, given that we know there is already one
36 Probability that you roll a 6 with a single roll of a fair die, given that the person before you rolled a 6 If the die is fair, this knowledge does not help. P(6 6 previously)=p(6)=1/6 Probability that you rolled a prime number {2,3,5} given that you rolled an odd number There are 3 odd numbers, {1,3,5}, but only {3,5} are prime P(prime odd)=2/3 Rules for Conditional Probabilities Conditional Probability For events A and B in a sample space S, the conditional probability of A given B is n(a B) n(a B) P(A B)= = n(s) P(A B) = n(b) n(b) P(B) n(s) P(B) 0
37 Intersection of Events Product Rule Last time we discussed the probability of a union of two events, what about intersection? P(A B) P(A B) P(A B)= and P(B A)= P(B) P(A) so P(A B)=P(A B)P(B)=P(B A)P(A) If 60% of the customers of a department store are female and 80% of the male customers have charge accounts, what is the probability that a customer selected at random is male and has a charge account? P(male)=1P(female)=10.6=0.4 P(charge male)=0.8 P(male and charge)=p(charge male)p(male) = 0.8 * 0.4 =0.32
38 Probability Trees A box has two green (striped) and three blue (dotted) balls. We draw two balls without replacement. g 2 S start g 1 b 1 b 2 g 2 b 2 S start P(g 2 g 1 )= g 1 b 1 P(b 2 b 1 )= g 2 b 2 g 2 b 2
39 P(g g 1 P(g b 1 P(b g 1 P(b 1 b )=P(g )P(g g )= )=P(g )P(b g )= )=P(b )P(g b )= ) = P(b )P(b b ) = = = = = 4 10 Total =1 Constructing Probability Trees Step 1. Draw tree diagram, all combined outcomes of sequences of experiments Step 2. Assign probabilities to each branch; probability of combined outcome is product of all probabilities leading to that end Step 3. Use the results of the above to answer all questions
40 More Draws, Bigger Tree Draw three times without replacement, what is the probability to draw a) striped, striped, dotted (g,g,b)=g 1 g 2 b 3 b) dotted, dotted, dotted (b,b,b)=b 1 b 2 b 3 c) striped, striped, striped (g,g,g)=g 1 g 2 g 3
41 g 2 b 3 g 3 g 1 b 2 S b 3 g 3 g 2 b 1 b 3 g 3 b 2 b 3 Independence Suppose instead of keeping the balls out after they were drawn, we put them back in, then it looks as follows: After 1 draw After 2 draws After 3 draws
42 Independence If A and B are any events then we say that A and B are independent if and only if P(A B) = P(A)P(B) Otherwise, And B are dependent Theorem 1: On Independence If A and B are independent events with nonzero probability, then P(A B)=P(A) and P(B A)=P(B) If either of the above equations holds, then A and B are independent.
43 Testing for Independence We need to check whether the probability that events A and B occur together is the same as the product of the probability of A and the probability of B Toss a coin twice: S={HH,HT,TH,TT} A={HH,HT} B={HH,TH} P(A and B)=P(HH)=1/4 P(A)=1/2; P(B)=1/2; P(A)P(B)=1/2*1/2=1/4 Independent of Set of Events A set of events is said to be independent if for any finite subset {E 1,E 2,,E k } we have P(E E E )=P(E ) P(E ) 1 2 k 1 2 P(E ) k
44 Problem 50 page 434 Quality control: A car manufacturer produces 37% of its cars at plant A. If 5% of the cars made at plant A have a defective part, what is the probability that car of this manufacturer was made at plant A and has a defective part? A=made at plant A; D=defective part P(A)=0.37, P(D A)=0.05 P(A and D)=0.37*0.05= Summary Conditional Probability P(A B)= P(A B) and P(B A)= P(A B) P(B) P(A) Product Rule P(A B)=P(A B)P(B)=P(B A)P(A) Independen t Events P(A B)=P(A)P(B) P(A B)=P(A) and P(B A)=P(B) if P(A) 0 and P(B) 0 If E,E,,E are independent then 1 2 k P(E E E )=P(E ) P(E ) P(E ) 1 2 k 1 2 k
45 Bayes Formula Probability of an earlier event g 2 g 1 S start b 2 g 2 b 1 b 2 g 2 g 1 S start b 2 g 2 b 1 b 2 Given the probabilities at the end P(b 2 ), P(g 2 ) can we determine P(g 1 ), P(b 1 )?
46 A more useful setting Given are two urns U 1, containing 3 blue and two white balls, and U 2, containing one blue and 3 white balls. We first pick an urn, then pick a ball at random from the chosen urn. P(U 1 )=1/3 P(B U 1 )=3/5 P(W U 1 )=2/5 P(U 2 )=2/3 P(B U 2 )=1/4 P(W U 2 )=3/4 The probability tree 3 5 B S start U 1 U W B 3 4 W
47 Now we would like to know: If we got a blue ball what was the probability that it came from urn 1? P(U1 B) P(U1 B) = P(B) Since there are only two ways to get B P(B) = P(U B) + P(U2 1 B) Combining these two P(U1 B) P(U1 B) = P(U B) + P(U 1 2 B) P(U 1)P(B U 1) P(U1 B) = P(U )P(B U ) + P(U )P(B U ) P(U 1 1 P(B U 1)P(U 1) B) = P(B U )P(U ) + P(B U )P(U 2 )
48 Recall from last time P(B U 1 )P(U 1 )=Product of branch probabilities leading to B through U 1 : (3/5)*(1/3)=1/5 P(B U 2 )P(U 2 )= Product of branch probabilities leading to B through U 2 : (1/4)*(2/3)=1/6 Product of branch probabilities to B through U1 P(U1 B) = Sum of all branch products to B = = More Urns, a Picture S E U 1 E U 2 E U 3 E U 1 U 2 U 3
49 Associated Formulae P(U1 E) P(U1 E) = P(U E) + P(U E) + P(U E) P(U 1 P(E U 1)P(U 1) E) = P(E U )P(U ) + P(E U )P(U ) + P(E U )P(U 3 ) Bayes Formula Theorem 1 Let U 1,U 2,, U n be n mutually exclusive events whose union is S. Let E be an event in S such that P(E) is not zero. Then P(U k P(Uk E) E) = P(E) P(U k P(E Uk )P(Uk ) E) = P(E U )P(U ) P(E U 1 1 n )P(U n )
50 3 5 B U 1 S start U 1 U W B S start B W U 2 U W U 2 Page 442, number 42: A company rated 75% of employees satisfactory, 25%unsatisfactory. Of the satisfactory ones 80% had experience, of the unsatisfactory only 40%. If a person with experience is hired, what is the probability that (s)he will be satisfactory? E=experience, N=no experience, S=satisfactory U=unsatisfactory P(E S)=0. 8, P(E U)=0.2, P(N S)=0.4, P(N U)=0.6 P(S)=0.75, P(U)=0.25 Question, what is P(S E)?
51 S U 0.6 E N E N P(E S)P(S) P(S E) = = P(E S)P(S) + P(E U)P(U) Random Variable Probability Distribution Expectation RANDOM VARIABLE, PROBABILITY DISTRIBUTION EXPECTED VALUE OF A RANDOM VARIABLE DECISIONMAKING AND EXPECTED VALUE
52 Random Variable A random variable is a function that assigns a numerical value to each simple event in a sample space S. Example: Tossing a coin. S 1 ={H,T} for a single toss S 2 ={HH,HT,TH,TT} two tosses The random variable will depend on the question that we want to answer: Number of heads: f(h)=1, f(t)=0 g(hh)=2, g(ht)= g(th)=1, g(tt)=0
53 Simpler notation: Tossing a coin three times, let x be the random variable of the number of tails. x { 0,1,2,3} The corresponding events are x = 0 if the event is {HHH} x = 1if the event is {HTT,THT,TTH} x = 2 if the event is {HHT,THH,HTH} x = 3 if the event is {HHH} What about probabilities? Probability of getting two tails when tossing a coin three times: x= number of tails p(0)=p(x=0)=p({e i in S x(e i )=0})=P({HHH})=1/8 p(1)=p(x=1)=p({e i in S x(e i )=1}) =P({THH,HTH,HHT})=3/8 p(2)= =3/8 p(3)= =P({TTT})=1/8
54 Probability distribution of a random variable X A probability function P(X=x)=p(x) is a probability distribution of the random variable X if 1. 0<p(x)<1, x in {x 1,x 2,,x n } 2. P(x 1 )+p(x 2 )+ +p(x n )=1 where {x 1,x 2,,x n } are the range (possible values) of X. Random variable X Probability distribution p e 1, e 2 e 3 e 4 e 5 e 6 e 7 e 8 x 1, x 2 x 3 x 4 x 5 x 6 p 1 p 2 p 3 Sample space Numerical values of X Probability of X=x
55 Expected Values of a Random Variable IDEA: What happens in the long run? What is the average number of tails in the experiment of tossing the coin three times? 1) toss three times 2) record number of tails {0,1,2,3} 3) repeat n times We expect to get 0 tails 1/8 of the time, 1 tail 3/8 of the time, 2 tails 3/8 of the time and 3 heads 1/8 of the time Expected value of a random variable X Given the probability distribution of a random variable X, x i x 1 x 2 x n p i p 1 p 2 p n Where p i = p(x i ), we define the expected value of X, denoted E(X), by the formula E(X)=x 1 p 1 +x 2 p 2 + +x n p n
56 Steps for Computing Expected Values Step 1: Form the probability distribution of the random variable X. Step 2: Multiply each image value of X by its corresponding probability of occurrence; add the results. Expected Winnings for Lottery Suppose we pay $ D (that is buy D tickets with the same six numbers) and win one million dollars if our six numbers come up, otherwise loose our $ D. What are our expected winnings? This is a 6 out of 49 lotto.
57 Correct combinations: ! 6! 43! Possible combinations: = = 13,983,816 Probability of winning: = Probability of loosing: = Expected winnings = (1,000,000D)* (D)*( ) = D Fair would be if expected value = 0 This would correspond to a ticket price of D= or about 7 cents If you bet $1 you can expect to win or loose about 93 cents
58 Decision making Outdoor concert, weather forecast predicts 0.24 chance of rain. If no rain $ net, if rain only $ net. Insurance for the $ costs $ Should you insure? Insured Not insured p i 0.24 rain 0.76 no rain x i x i $ 90,000 $ 10,000 $ 80,000 $100,000 Ins. E(X)=(90000)(0.24)+(80000)(0.76)=82400 Not E(X)=(10000)(0.24)+(100000)(0.76)=78400
59 The other side of the story: For the insurance company: E(X)=(80000)(0.24)+(20000)(0.76)=4000 They are either soon going out of business or have inside information.
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