Math 5. Rumbos Fall 23 Solutions to Review Problems for Final Exam. Three cards are in a bag. One card is red on both sides. Another card is white on both sides. The third card in red on one side and white on the other side. A card is picked at random and placed on a table. Compute the probability that if a given color is shown on top, the color on the other side is the same as that of the top. Solution: Each card has a likelihood of /3 of bing picked. Assume for definiteness that the top of the picked card is red. Let T r denote the event that the top of picked car shows red and B r denote the event that the bottom of the card is also red. We want to compute Pr(B r T r ) = Pr(T r B r ). () Pr(T r ) Note that Pr(T r B r ) = 3, (2) since there is only one card for which both sides are red. In order to compute Pr(T r ) observe that there are three equally likely choices out of six for the top of the card to show red; thus, Hence, using (2) and (3), we obtain from () that Pr(T r ) = 2. (3) Pr(B r T r ) = 2 3. (4) Similar calculations can be used to show that Pr(T w ) = 2, (5) and Pr(B w T w ) = 2 3. (6) Let E denote the event that a card showing a given color on the top side will have the same color on the bottom side. Then, by the law of total probability, Pr(E) = Pr(T r ) Pr(B r T r ) + Pr(T w ) Pr(B w T w ), (7)
Math 5. Rumbos Fall 23 2 so that, using (2), (4), (5) and (6), we obtain from (7) that Pr(E) = 2 2 3 + 2 2 3 = 2 3. 2. An urn contains balls: 4 red and 6 blue. A second urn contains 6 red balls and a number b of blue balls. A single ball is drawn from each urn. The probability that both balls are the same color is.44. Determine the value of b. Solution: Let R denote the event that the ball drawn from the first urn is red, B denote the event that the ball drawn from the first urn is blue, R 2 denote the event that the ball drawn from the second urn is red, and B 2 denote the event that the ball drawn from the second urn is blue. We are interested in the event E = (R R 2 ) (B B 2 ), the event that both balls are of the same color. We observe that R R 2 and B B 2 mutually exclusive events; thus, Pr(E) = Pr(R R 2 ) + Pr(B B 2 ), (8) where R and R 2 are independent events as well as B and B 2. It the follows from (8) that where Pr(E) = Pr(R ) Pr(R 2 ) + Pr(B ) Pr(B 2 ), (9) Pr(R ) = 4, Pr(R 2) = 6 6 + b, Pr(B ) = 6, and Pr(B 2) = b 6 + b ; thus, substituting into (9), Pr(E) = 4 6 6 + b + 6 b 6 + b. () We are given that Pr(E) =.44; combining this information with () yields 32 + 3b 6 + b = 2.2. () Solving () for b yields b = 4.
Math 5. Rumbos Fall 23 3 3. A blood test indicates the presence of a particular disease 95% of the time when the disease is actually present. The same test indicates the presence of the disease.5% of the time when the disease is not present. One percent of the population actually has the disease. Calculate the probability that a person has the disease given that the test indicates the presence of the disease. Solution: Let D denote that a person selected at random from the population has the disease. Then, Pr(D) =.. (2) Let P denote the event that the blood test is positive for the existence of the disease. We are given that and We want to compute where by the Multiplication Rule, so that by virtue of (2) and (3), and Pr(P D) =.95 (3) Pr(P D c ) =.5. (4) Pr(D P ) = Pr(D P ), (5) Pr(P ) Pr(D P ) = Pr(D) Pr(P D), Pr(D P ) =.95, (6) Pr(P ) = Pr(D) Pr(P D) + Pr(D c ) Pr(P D c ), (7) by the Law of Total Probability. Substituting the values in (2), (3) and (4) into (7) yields Pr(P ) = (.) (.95) + (.99) (.5) =.445. (8) Substituting the values in (6) and (8) into (5) yields Pr(D P ) =.6574. Thus, it a person tests positive, there is about a 66% chance that she or he has the disease.
Math 5. Rumbos Fall 23 4 4. A study is being conducted in which the health of two independent groups of ten policyholders is being monitored over a one-year period of time. Individual participants in the study drop out before the end of the study with probability.2 (independently of the other participants). What is the probability that at least 9 participants complete the study in one of the two groups, but not in both groups? Solution: Let X denote the number of participants in the first group that drop out of the study and Y denote the number of participants in the second group that drop out of the study. Then X and Y are independent Binomial(,.2) random variables. Then, A = (X ) is the event that at least 9 participants complete the study in the first group, and B = (B ) is the event that at least 9 participants complete the study in the second group. We cant to compute the probability of the event so that E = (A B c ) (A c B), Pr(E) = Pr(A B c ) + Pr(A c B), (9) since A B c and A c B are mutually exclusive. Next, use the independence of A and B, given that X and Y are independent random variables, to get from (9) that Pr(E) = Pr(A) ( Pr(B) + Pr(A) ( Pr(B)), (2) where Pr(A) = Pr(X ) so that Similarly, = Pr(X = ) + Pr(X = ) = (.8) + (.2) (.8) 9, Pr(A) =.3758. (2) Pr(B) =.3758. (22) Substituting the results in (2) and (22) into (2) yields Pr(E) =.469. Thus, the probability that at least 9 participants complete the study in one of the two groups, but not in both groups, is about 47%.
Math 5. Rumbos Fall 23 5 5. Suppose that < ρ < and let p(n) = ρ n ( ρ) for n =,, 2, 3,... (a) Verify that p is the probability mass function (pmf) for a random variable. Solution: Compute ρ n ( ρ) = ( ρ) ρ n = ( ρ) ρ =, n= n= since < ρ < and, therefore, the geometric series n= ρ n converges to ρ. (b) Let X denote a discrete random variable with pmf p. Compute Pr(X > ). Solution: Compute Pr(X > ) = Pr(X ) = p() p() = ( ρ) ρ( ρ) = ρ 2. 6. In modeling the number of claims filed by an individual under an automobile policy during a three-year period, an actuary makes the simplifying assumption that for all integers n, p n+ = 5 p n, that the policyholder files n claims during the period. where p n represents the probability (a) Find the value of p o for which p n defines the probability mass function for the number, N, of claims filed during that period. Solution: We first determine the values of p n by solving p n+ = 5 p n recursively: p = 5 p o p 2 = ( ) 2 5 p = p o 5 p 3 = ( ) 3 5 p 2 = p o 5.
Math 5. Rumbos Fall 23 6 so that n= p n = ( ) n p o, for n =,, 2,... (23) 5 The expression in (23) can also be established by induction on n. Next, compute ( ) n p n = p o = p o 5 (/5) = 5 4 p o; n= thus, in order for p n to define a probability mass function, it must be the case that p o = 4 5. (24) (b) What is the probability that a policyholder files more than one claim during the period? Solution: Compute Thus, using (23) and (24), Pr(N > ) = Pr(N ) = Pr(N = ) Pr(N = ) = p o p. Pr(N > ) = 4 5 5 4 5 = 25. Hence, he probability that a policyholder files more than one claim during the period is 4%. 7. The time, T, that a manufacturing system is out of operation has cumulative distribution function ( ) 2 2, for t > 2; t F T (t) =, elsewhere. The resulting cost to the company is proportional to Y = T 2. Determine the probability density function for Y.
Math 5. Rumbos Fall 23 7 Solution: Firs, compute the cdf of Y = T 2 to get F Y (y) = Pr(Y y), for y > 4; = Pr(T 2 y) = Pr(T y) = F T ( y), so that 4, for y > 4; y F Y (y) =, elsewhere. It then follows from (25) that the pdf for Y is 4, for y > 4; y2 f Y (y) =, elsewhere. (25) 8. Let N(t) denote the number of mutations in a bacterial colony that occur during the interval [, t). Assume that N(t) Poisson(λt) where λ > is a positive parameter. (a) Give an interpretation for λ. Answer: λ is the average number of mutations per unit of time. To see why this assertion makes sense, note that, since N(t) Poisson(λt), E(N(t)) = λt, for t >. Thus, λ = E(N(t)), for t >. t (b) Let T denote the time that the first mutation occurs. Find the distribution of T.
Math 5. Rumbos Fall 23 8 Solution: Observe that, for t >, the event [T > t] is the same as the event [N(t) = ]; that is, if t < T, there have have been no mutations in the time interval [, t]. Consequently, since N(t) Poisson(λt). Thus, Pr[T > t] = Pr[N(t) = ] = e λt, Pr[T t] = Pr[T > t] = e λt, for t >. We then have that the cdf of T is { e λt, for t > ; F T (t) = for t. (26) It follows from (26) that the pdf for T is { λe λt, for t > ; f T (t) = for t, which is the pdf for an exponential distribution with parameter β = /λ; thus, T Exponential(/λ). 9. Let X Exponential(β), for β >. Compute the median of X. Solution: The median of the distribution of a continuous random variable X is the value of x, denoted by m, for which in other words, the value m for which Pr(X m) = 2 ; F X (m) = 2. (27) The cdf for X Exponential(β) is given by { e x/β, for x > ; F X (x) = for x. (28)
Math 5. Rumbos Fall 23 9 Thus, in view of (27) and (28), we see that the median of the distribution of X Exponential(β) is a positive value, m, such that e m/β = 2. (29) Solving (29) for m yields m = β ln(2).. Two checkers at a service station complete checkouts independent of one another in times T Exponential(µ ) and T 2 Exponential(µ 2 ), respectively. That is, one checker serves /µ customers per unit time on average, while the other serves /µ 2 customers per unit time on average. (a) Give the joint pdf, f T,T 2 (t, t 2 ), of T and T 2. Solution: Since T and T 2 are independent random variables, the joint pdf of (T, T 2 ) is given by f (T,T 2 ) (x, y) = f T (x) f T (y), for (x, y) R 2, (3) where e x/µ, for x > ; µ f T (x) = for x, and e y/µ 2, for y > ; µ 2 f T2 (y) = for y. It follows from (3), (3) and (32) that the joint pdf of (T, T 2 ) is e t /µ t 2 /µ 2, for t > and t 2 > ; f (t µ µ 2 (T,T 2 ), t 2 ) = elsewhere. (3) (32)
Math 5. Rumbos Fall 23 (b) Define the minimum service time, T m, to be T m = min{t, T 2 }. Determine the type of distribution that T m has and give its pdf, f Tm (t). Solution: Observe that, for t >, the event [T m > t] is the same as the event [T > t, T 2 > t], since T m is the smaller of T and T 2. Consequently, Pr[T m > t] = Pr[T > t, T 2 > t]. (33) Thus, by the independence of T and T 2, it follows from (33) that where so that Pr[T m > t] = Pr[T > t] Pr[T 2 > t], (34) Pr[T i > t] = F Ti (t) = ( e t/µ i ), Pr[T i > t] = e t/µ i, for t > and i =, 2. (35) Combining (34) and (35) then yields or where we have set Pr[T m > t] = e t/mu t/µ 2, Pr[T m > t] = e t/β, (36) β = µ + µ 2. (37) It follows from (36) that the cdf of T m is { e t/β, for t > ; F Tm (t) = for t, where β is given by (37). Thus, the pdf for T m is β e t/β, for t > ; f Tm (t) = for t, (38) which is the pdf for an exponential distribution with parameter β given by (37); thus, T m Exponential(β), where β = µ µ 2 µ + µ 2. (39)
Math 5. Rumbos Fall 23 t 2 L R L t Figure : Region for Problem (c) Suppose that, on average, one of the checkers serves 4 customers in an hour, and the other serves 6 customers per hour. On average, what is the minimum amount of time that a customer will spend being served at the service station? Solution: We compute the expected value of T m, where T m has pdf given in (38) with β = 4 6 4 + =, 6 in view of (39). Thus, on average, the minimum time spent by a customer being served at the service station is one tenth of an hour, or 6 minutes.. Let T and T 2 represent the lifetimes in hours of two linked components in an electronic device. The joint density function for T and T 2 is uniform over the region defined by t t 2 L, where L is a positive constant. Determine the expected value of the sum of the squares of T and T 2. Solution: The region R = {(t, t 2 ) R 2 t t 2 L} is pictured in Figure. The area of R is area(r) = L2 2. Thus, since (T, T 2 ) has a uniform distribution over R, it follows that the joint
Math 5. Rumbos Fall 23 2 pdf of T and T 2 is given by 2 f (t L, for t (T,T 2 ), t 2 ) = 2 t 2 L;, elsewhere. We want to compute E(T 2 + T2 2 ), or E(T 2 + T2 2 ) = (t 2 + t 2 2)f (t (T R 2,T 2 ), t 2 ) dt dt 2 = R = 2 L 2 L (t 2 + t 2 2) 2 L 2 dt dt 2 = 2 L 2 L = 2 L 2 L L = 2 L [ L 3 L 2 t (t 2 + t 2 2) dt 2 dt [ ] L (t 2 t 2 + t3 2 3 ) dt t ( )] [Lt 2 + L3 3 t 3 + t3 3 3 + Lt2 4 3 t3 ] dt dt so that = 2 L 2 [ L 3 3 t + 3 Lt3 3 t4 E(T 2 + T 2 2 ) = 2 3 L2. ] L, 2. A device runs until either of two components fails, at which point the device stops running. The joint density function of the lifetimes of the two components, both measured in hours, is f(x, y) = x + y, for < x < 2 and < y < 2; 8
Math 5. Rumbos Fall 23 3 y 2 2 x Figure 2: Region for Problem 2 and elsewhere. What is the probability that the device fails during its first hour of operation? Solution: We want to compute the probability that either component fails within the first hour of operation; that is the probability of the event E given by E = ( < X < ) ( < Y < ). The event E is pictured as the shaded region in Figure 2. The probability of E is given by Pr(E) = f(x, y) dxdy = = 8 E 2 ( x + y 8 dydx + ] 2 [xy + y2 2 = ( (2x + 2) dx + 8 = 8 2 dx + 2 ( [x 2 + 2x ] + [ x 2 2 + x 2 so that Pr(E) = 8 (3 + (3 )) = 5 8. fails during its first hour of operation is 62.5%. 2 x + y 8 dydx ] [xy + y2 2 ( x + ) 2 ] 2 ), ) dx dx Thus, the probability that the device )
Math 5. Rumbos Fall 23 4 3. A device that continuously measures and records seismic activity is placed in a remote region. The time, T, to failure of this device is exponentially distributed with mean 3 years. Since the device will not be monitored during its first two years of service, the time to discovery of its failure is X = max(t, 2). Determine E[X]. Solution: The pdf of T is 3 e t/3, for t > ; f T (t) = for t, (4) where t is measured in years. The random variable X = max(t, 2) is also given by { 2, if T 2; X(T ) = T, if T > 2. Thus, the expected value of X(T ) is given by E(X(T )) = X(t)f T (t) dt, where f T is as given in (4) and X(t) = { 2, if t 2; t, if t > 2. We then have that E(X(T )) = 2 2 3 e t/3 dt + 2 t 3 e t/3 dt = 2 [ e t/3] 2 + [ te t/3 3e t/3], 2 where we have used integration by parts to integrate the last integral. We then have that E(X(T )) = 2 ( e 2/3) + 2e 2/3 + 3e 2/3 = 2 + 3e 2/3.
Math 5. Rumbos Fall 23 5 4. The number, Y, of spam messages sent to a server in a day has a Poisson distribution with parameter λ = 2. Each spam message independently has a probability p = /3 of not being detected by the spam filter. Let X denote the number of spam massages getting through the filter. Calculate the expected daily number of spam messages which get into the server. Solution: We are given that where λ = 2, so that Y Poisson(λ), (4) Pr[Y = n] = λn n! e λ, for n =,, 2,... Now, since each spam message independently has a probability p = /3 of not being detected by the spam filter, it follows that X has a conditional distribution (conditioned on Y = n) that is Binomial(n, p), with p = /3; thus, ( ) n p k ( p) n k, for k =,, 2,..., n; k Pr[X = k Y = n] = (42) elsewhere. Then, Pr[X = k] = Pr[X = k, Y = n] n= = Pr[Y = n] Pr[X = k Y = n], n= where Pr[X = k Y = n] = for n < k, so that, using (4) and (42), ( ) λ n n Pr[X = k] = n! e λ p k ( p) n k k n=k = e λ k! p k n=k λ n (n k)! ( p)n k. (43) Next, make the change of variables l = n k in the last summation in (43) to get Pr[X = k] = e λ k! p k l= λ l+k l! ( p)l,
Math 5. Rumbos Fall 23 6 so that Pr[X = k] = (λp)k k! = (λp)k k! e λ l= e λ e λ( p) [λ( p)]l l! = (λp)k e λp, k! which shows that X Poisson(λp), or X Poisson(7) We then have that E(X) = 7, so that the expected daily number of spam messages which get into the server is 7. 5. Two instruments are used to measure the height, h, of a tower. The error made by the less accurate instrument is normally distributed with mean and standard deviation.56h. The error made by the more accurate instrument is normally distributed with mean and standard deviation.44h. Assuming the two measurements are independent random variables, what is the probability that their average value is within.5h of the height of the tower? Solution: Denote the measurements by X and X 2 so that where X = h + E and X 2 = h + E 2, (44) E Normal(, (.56h) 2 ) and E 2 Normal(, (.44h) 2 ). (45) The average, X, of the two measurements is where we have used (44). Setting we can rewrite (46) as X = h + E + E 2, (46) 2 E = E + E 2 2 (47) X = h + E. (48) It follows from (47), (45) and the independence of E and E 2 that E Normal(, σ 2 ), (49)
Math 5. Rumbos Fall 23 7 where σ 2 = 4 [(.56h)2 + (.44h) 2 ] =.268h 2, so that We want to compute the probability or, in view of (48), σ =.356h (5) Pr( X h <.5h), Pr( E <.5h). (5) Dividing both sides of the inequality in (5) by σ in (5), and using the result in (49), we obtain from (5) that where Z Normal(, ). It follows from (52) that so that Pr( X h <.5h) = Pr( Z <.4), (52) Pr( X h <.5h) = F Z (.4) F Z (.4) = 2F Z (.4) ; Pr( X h <.5h) = 2(.992) =.8384. Thus, the probability that the average of the two measurements is within.5h of the height of the tower is about 84%. 6. The lifetime of a printer costing $2 is exponentially distributed with mean 2 years. The manufacturer agrees to pay a full refund to a buyer if the printer fails during the first year following its purchase, and a one-half refund if it fails during the second year. If the manufacturer sells printers, how much should it expect to pay in refunds? Solution: Let T denote the lifetime of a printer; then, the pdf of T is 2 e t/2, for t > ; f T (t) = for t, where t is measured in years. (53)
Math 5. Rumbos Fall 23 8 The expected cost on the sale of one printer is E(C) = 2 Pr( < T ) + Pr( < T 2), (54) where, using the pdf in (53), and Pr( < T ) = Pr( < T 2) = 2 2 e t/2 dt = e /2, (55) 2 e t/2 dt = [ e t/2] 2 = e /2 e. (56) Substituting the results in (55) and (56) into (54) yields E(C) = 2 e e /2 = 2.56. Thus, the expected payments because of returns on the sale of printers is about $, 256. 7. A company manufactures a brand of light bulb with a lifetime in months that is normally distributed with mean 3 and variance. A consumer buys a number of these bulbs with the intention of replacing them successively as they burn out. The light bulbs have independent lifetimes. What is the smallest number of bulbs to be purchased so that the succession of light bulbs produces light for at least 4 months with probability at least.9772? Solution: Denote the lifetimes of the light bulbs by T, T 2, T 3,..., so that T i Normal(3, ), for i =, 2, 3,..., are independent random variables measured in months. The total duration of n n light bulbs is T k. We want to find the smallest n so that k= ( n ) Pr T k > 4.9772. (57) k= Since the T i s are independent, normally distributed, it follows that n T k Normal(3n, n), k=
Math 5. Rumbos Fall 23 9 it then follows that n T k 3n k= n Normal(, ). (58) Using the information in (58), we can write the estimate in (57) as ( ) 4 3n Pr Z >.9772, n where Z Normal(, ), or F Z ( 4 3n n ).9772, form which we get ( ) 3n 4 F Z.9772. (59) n Using the symmetry for the pdf of standard normal distribution we can rewrite (59) as F Z ( 3n 4 n ).9772. (6) Looking at a table of standard normal probabilities, we see that (6) is satisfied if 3n 4 2. (6) n To find the smallest value of n for which (6) is satisfied, we solve the inequality in (6) to get n 6. Thus, the smallest number of bulbs to be purchased so that the succession of light bulbs produces light for at least 4 months with probability at least.9772 is 6. 8. A computer manufacturing company conducts acceptance sampling for incoming computer chips. After receiving a huge shipment of computer chips, the company randomly selects 8 chips. If three or fewer nonconforming chips are found, the entire lot is accepted without inspecting the remaining chips in the lot. If four or more chips are nonconforming, every chip in the entire lot is carefully inspected at the supplier s expense. Assume that the true proportion of nonconforming computer chips being supplied is.. Estimate the probability the lot will be accepted.
Math 5. Rumbos Fall 23 2 Solution: Let X denote the number of nonconforming chips found in the random sample of 8. We may assume that the tests of the chips are independent trials. Thus, X Binomial(n, p), where n = 8 and p =.. We want to estimate Pr(X 3). Since n is large and p is very small, we may use the Poisson approximation to the binomial distribution to get Pr(X 3) Pr(Y 3), where Y Poisson(.8). It then follows that Pr(X 3) e.8 + (.8)e.8 + (.8)2 2 e.8 + (.8)3 e.8 =.999; 6 thus, the probability the lot will be accepted is about 99.9%. 9. Last month your company sold, new watches. Past experience indicates that the probability that a new watch will need repair during its warranty period is.2. Estimate the probability that no more than 5 watches will need warranty work. Explain the reasoning leading to your estimate. Solution: Let X denote the number of watches out of the, that will need repair. We may assume that w given watch will needs repair is independent of the event that any watch in the batch will need repair. Thus, X Binomial(n, p), where n =, and p =.2. We want to estimate Pr(X 5). Since n is very large and p is very small, we may use the Poisson approximation to the binomial distribution to get Pr(X 5) Pr(Y 5), where Y Poisson(2). It then follows that Pr(X 3) 5 (2) k e 2 ; k! thus, the probability that no more than 5 watches will need warranty work is about 7.9 5, a very small probability. k=
Math 5. Rumbos Fall 23 2 y x = y Figure 3: Region for Problem 2 x 2. A device contains two circuits. The second circuit is a backup for the first, so the second is used only when the first has failed. The device fails when and only when the second circuit fails. Let X and Y be the times at which the first and second circuits fail, respectively. Assume that X and Y have joint probability density function { 6e x e 2y, for < x < y < ; f (X,Y ) (x, y) =, elsewhere. What is the expected time at which the device fails? Solution: First, compute the marginal distribution of Y, the time at which the second circuit fails; see Figure 3 to identify the region of integration. f Y (y) = = y f (X,Y ) (x, y) dx 6e x e 2y dx = 6e 2y [ e x] y = 6e 2y [ e y],
Math 5. Rumbos Fall 23 22 Thus, f Y (y) = Then, the expected value of Y is given by { 6(e 2y e 3y ), for y > ;, elsewhere. E(Y ) = yf Y (y) dy = 3 = 3 2 2 3 = 5 6. y 2e y/(/2) dy 2 y 3e y/(/3) dy