1. Suppose that we ask n randomly selected people whether they share your birthday. (a) Give an expression for the probability that no one shares your birthday (ignore leap years). (5 marks) Solution: For a randomly chosen person, the chanced that the person does not share your birthday is 1-(1/365). Thus for n randomly chosen people the chance is ( ) n 364 365 (b) How many people do we need to select so that the probability that at least two people share your birthday is greater than.5? (5 marks) Solution: Define A = {At least two people out of n share your birthday} and B k = {Exactly k people out of n share your birthday}, for k =, 1,, n. Then A = n k=2 B k and hence P (A) = n k=2 P (B k). Define { 1, if the i-th person share your birthday X i =, otherwise, and Y = n X iid i. Note that X i Bernoulli(p = 1 ) and hence 365 Y Binomial(p = 1, n). We also note that 365 ( ) ( ) k ( ) n k n 1 364 P (B k ) = P (Y = k) = k 365 365 and therefore P (A) = P (Y 2) = 1 P (Y < 2) = 1 P (Y = ) P (Y = 1) ( ) n ( ) 364 364 n 1 = 1 n 365 365 n = 1 364n 1 (364 + n) 365 n 1
2. A heavy-equipment salesperson can contact either one or two customers per day with probability 1/3 and 2/3, respectively. Each contact will result in either no sale or a $5, sale, with the probabilities.9 and.1, respectively. Give the probability distribution for daily sales. Find the mean and standard deviation of the daily sales. (1 marks) Solution: Let X be the number of customers who are contacted on a randomly chosen day. Suppose W i is the sale resulted from the i-th contact. Then the daily sale is S = X W i. Therefore, E(S) = E[E(S X)] = E[ X E(W i )] = E(X)µ W where µ W = E(W i ) = (.9) + 5, (.1) = 5, for all i s. On the other hand, E(X) = 1 1/3 + 2 2/3 = 5/3 and therefore E(S) = (5/3)5, = 25, /3. To find the variance we have V(S) = E[V(S X)] + V[E(S X)] = E[XV(W i )] + V[XE(W i )] = µ X σ 2 W + σ 2 Xµ 2 W. We calculated µ X and µ W above. It remains to calculate the variances. To do so, we need to find the second moments of X and W. and hence E(X 2 ) = 1 2 1/3 + 2 2 2/3 = 9/3 = 3 E(W 2 ) = 2 (.9) + (5, ) 2 (.1) = 25 1 7, σ 2 X = V(X) = E(X 2 ) [E(X)] 2 = 3 (5/3) 2 = 2/9 σ 2 W = V(W ) = E(W 2 ) [E(W )] 2 = 25 1 7 [5] 2 = 225 1 6. We therefore have σ S = V(S) = µ X σ 2 W + σ2 X µ2 W = (5/3)(225 1 6 ) + (2/9)(25 1 6 ) 137 = 25 16 9 3425 = 1 9 6 1957.83 2
3. The length of time necessary to complete a key operation in the construction of houses has an exponential distribution with mean 1 hours. The formula C = 1 + 4Y + 3Y 2 relates the cost C of completing this operation to Y. Find the mean and variance of C. (1 marks) Solution: We have Y Exp(1/1). To find the mean and variance of C we need to find the first four moments of C. E(Y k ) = y k 1 β e y/β dy = 1 β y k e y/β dy. Note that the last integral in the above expression is a Γ integral. Recall the following identity which follows immediately from the fact that Γ density integrates to 1, Using the above identity, we have y α 1 e y/β = Γ(α)β α. y k e y/β dy = Γ(k + 1)β k+1 = k!β k+1, where the last equality follows from the fact that Γ(α + 1) = αγ(α) for all α >. Therefore E(Y k ) = 1 β k!βk+1 = k!β k. It follows from properties of Expectation operator that E(C) = 1+4E(Y )+3E(Y 2 ) = 1+4β+3(2!β 2 ) = 1+4(1)+6(1) = 11. To find the variance we first find E(C 2 ), E(C 2 ) = 1 4 + 16E(Y 2 ) + 9E(Y 4 ) + 4E(Y ) + 3E(Y 2 ) + 12E(Y 3 ) Therefore = 1 4 + 4 1 3 (1!β 1 ) + 19 1 2 (2!β 2 ) + 12(3!β 3 ) + 9(4!β 4 ) = 1 4 + 4 1 4 + 38 1 4 + 72 1 4 + 216 1 4 = 331 1 4. V(C) = E(C 2 ) [E(C)] 2 = 331 1 4 [11] 2 = 21 1 5. 3
4. The number of calls coming into a police emergency center in the time interval (, t] follows a Poisson distribution with mean λt. Let T denote the length of time until the first arrival time. (a) Find the probability density function of T. (5 marks) Solution: Let N(t) denote the number of calls coming into the police emergency center in the time interval [, t]. Then for any t > we have F T (t) = 1 P (T > t) = 1 P (N(t) = ) = 1 e λt (λt)! = 1 e λt, and hence f T (t) = df T (t)/dt = λe λt for t > and otherwise. In other words, T Exp(1/λ). (b) If the calls arrives at the rate of 1 calls per hour, what is the probability that more than 15 minutes elapse until the first call arrive? (5 marks) Solution: Note that λ = 1. We need to calculate P (T > 15/6). P (T > 1/4) = 1/4 1e 1y dy = e 1/4 = e 2.5. 4
5. A machine used to fill cereal boxes dispenses, on the average, µ ounces per box. the manufacturer wants the actual ounces dispensed, Y, to be within 1 ounce of µ at least 75% of the time. What is the largest value of σ, the standard deviation of Y, that can be tolerated if the manufacturer s objectives are to be met? (1 marks) Solution: Using Tchebyshev s inequality we have P ( Y µ Y < 2σ Y ) 3 4. The bound need to be 1. Thus 2σ Y = 1 and hence σ Y should be at most 1/2. 5
6. Suppose that the random variables Y 1 and Y 2 have joint probability density function, f(y 1, y 2 ), given by { 6(1 y 2 ), if y 1 y 2 1, f(y 1, y 2 ) =, otherwise. (a) Find P (Y 2 1/2 Y 1 3/4). (5 marks) Solution: We first find P (Y 1 3/4). We start with pdf of Y 1. For any y 1 1 we have f Y1 (y 1 ) = We therefore find On the other hand, f Y1,Y 2 (y 1, y 2 )dy 1 = P (Y 1 3/4) = P (Y 2 1/2, Y 1 3/4) = We can now easily find 3/4 1/2 y2 1 y 1 6(1 y 2 )dy 2 = 3(1 y 1 ) 2. 3(1 y 1 ) 2 dy 1 = 63 64. 6(1 y 2 )dy 1 dy 2 = 6 1/2 = 6[ 1 2 y2 2 1/2 1 3 y3 2 1/2 ] = 6[ 1 8 1 24 ] = 1/2. y 2 (1 y 2 )dy 2 P (Y 2 1/2 Y 1 3/4) = P (Y 2 1/2, Y 1 3/4) P (Y 1 3/4) = 1/2 63/64 = 32 63. (b) Find P (Y 2 3/4 Y 1 = 1/2). (5 marks) Solution: Note that f Y2 Y 1 (y 2 y 1 = 1/2) = f Y 1,Y 2 (y 1, y 2 ) f Y1 (1/2) = { 6(1 y2 ) 3[1 (1/2)] 2, if 1/2 y 2 1, otherwise. Thus, P (Y 2 3/4 Y 1 = 1/2) = 1 f Y2 Y 1 (y 2 y 1 = 1/2)dy 2 = 1 3/4 3/4 8(1 y 2 )dy 2 = 1 4. 6
7. A large lot of manufactured items consists of 1% items with exactly one defect, 5% with more than one defect and the remainder with no defects. Ten items are randomly selected from this lot for sale. If Y 1 denotes the number of items with one defect and Y 2, the number with more than one defect, the repair costs are Y 1 + Y 2. Find the mean and variance of the repair costs. (1 marks) Solution: Define Y 3 = 1 (Y 1 + Y 2 ) and note that (Y 1, Y 2, Y 3 ) Multinomial(n = 1, p 1 =.1, p 2 =.5, p 3 =.85). As such E(Y i ) = np i, V(Y i ) = np i (1 p i ) for i = 1, 2, 3 and Cov(Y i, Y j ) = np i p j, for i j. Then while E(Y 1 + Y 2 ) = E(Y 1 ) + E(Y 2 ) = 1 (.1) + 1 (.5) = 1.5, V(Y 1 + Y 2 ) = V(Y 1 ) + V(Y 2 ) + 2Cov(Y 1, Y 2 ) = 1(.1)(.9) + 1 (.5)(.95) 2 1(.1)(.5) = 1.275 7
8. The manager of a construction job needs to figure prices carefully before submitting a bid. He also needs to account for uncertainty (variability) in the amounts of products he might need. To simplify the real situation, suppose that a project manager treats the amount of sand, in yards, needed for a construction project as a random variable Y 1, which is normally distributed with mean 1 yards and standard deviation.5 yard. The amount of cement mix needed, in hundreds of pounds, is a random variable Y 2, which is normally distributed with mean 4 and standard deviation.2. The sand cost $7 per yard, and the cement mix costs $3 per hundred pounds. Adding $1 for other costs, he computes his total cost to be U = 1 + 7Y 1 + 3Y 2. If Y 1 and Y 2 are independent, how much should the manager bid to ensure that the true costs will exceed the amount bid with a probability of only.1? (1 marks) Solution: We first show that U is normally distributed. Let V = k a iy i where Y i s are independent random variable and Y i N(µ i, σ 2 i ) for i = 1, 2,, k, and also a 1,, a k are real numbers. Then which shows that m V (t) = E(e tv ) = E( = k k e ta iy i ) = e a iµ i t+ a 2 i σ2 i t2 2 = exp{t V N( Likewise, we can easily show that a i µ i, U = V + C N(C + where C is a constant. Thus k m Yi (a i t) a i µ i + t2 2 a 2 i σi 2 ). a i µ i, a 2 i σi 2 }. a 2 i σi 2 ), U = 1+7Y 1 +3Y 2 N(1+7 1+3 4, 7 2 (.5) 2 +3 2 (.2) 2 ) = N(182, 12.61) Having found the distribution of U, we need to find b such that.1 = P (U > b) = P (z > b 182 12.61 ). Using the attached table for normal distribution we find that and hence b = 211.38. b 182 12.61 = 2.33, 8
9. The length of time required for the periodic maintenance of an automobile or another machine usually has a mound-shaped probability distribution. Because some occasional long service times will occur, the distribution tends to be skewed to the right. Suppose that the length of time required to run a 5-mile check and to service an automobile has mean 1.4 hours and standard deviation.7 hour. Suppose also that the service department plans to service 5 automobiles per 8-hour day and that, in order to do so, it can spend a maximum average service time of only 1.6 hours per automobile. On what proportion of all workdays will the service department have to work overtime?(1 marks) Solution: Since the maximum average service time is 1.6, the job should be finished at most in 1 days. Note that 1 8 = 5 1.6. We need to see what is the chance that the job is finished in 8 hours. Let Y i be the time required to repair the i-th machine for i = 1, 2,, 5. Then we need to find P ( 5 i Y i > 8). Using CLT, we have 5 P ( Y i > 8) P (Z > i 8 5 1.4 5.7 ) = P (Z > 2.2) and we find from the attached table that P (Z > 2.2) =.2. In other words, on the average almost 2% of the machines require more than 1.6 hour. This then means that on the average 2% of the workdays the service department have to work overtime. 9
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STUDENT NAME: STUDENT ID# McGILL UNIVERSITY FACULTY OF SCIENCE FINAL EXAMINATION MATH323 PROBABILITY THEORY Examiner: Professor M. Asgharian Date: Wednesday, December 16, 29 Associate Examiner: Professor W. Anderson Time: 14: AM - 17: AM. INSTRUCTIONS Answer ONLY 8 questions. Calculators are permitted. Answer directly on the exam. Dictionaries are allowed. This is a closed book exam. Questions 1 2 3 4 5 6 7 8 9 Marks This exam comprises the cover, 9 pages of questions, and page 1 which is blank, 2 pages of formulas and one table.