Problem 1. Problem 2. Problem 3. Problem 4

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1 Problem Let A be the event that the fungus is present, and B the event that the staph-bacteria is present. We have P A = 4, P B = 9, P B A =. We wish to find P AB, to do this we use the multiplication rule, page 46. This is the st option, X% chose this. Problem P AB = P B AP A = Let X be a random variable describing the time it takes for the signal to be processed by a satelite. We have EX = s, V arx = 6s, and wish to find an upper bound for P X 5. To do this, we use Chebychev s inequality, page 9. We find This is the nd option, X% answered this. Problem P [ X EX ksdx] k P P [ X 6 V arx] 6 Let X be a random variable describing the number of ball bearings passing quality control. We have X Binom 4, 4 5, and wish to find SDX. We find the variance of a binomial distribution on page 477, inserting here we find This is the 5th option, X% answered this. Problem 4 SDX = np p = = 8 Let X, Y be random variables following a standard normal distribution describing the coordinates of a laser s offset from the center. We wish to find P X + Y, and as shown on page 59 the random variable Z = X + Y will follow a Rayleigh distribution. We can thus find the desired probability by looking up the cdf. of the Rayleigh distribution on page 478. This is the 5th option, X% chose this. P Z = P Z = e = e

2 Problem 5 Given the standardized random variables X, Y we know that EXY =. We wish to find V arx Y, to do this we start by finding the covariance CovX, Y, by the formula on page 4 we have CovX, Y = EXY EXEY =. By the formula on the same page we then find the variance This is the st option, X% answered this. Problem 6 V arx Y = V arx + V ary CovX, Y = + = Given an outcome space Ω the events B i, i =,, satisfy B i B j = for i j and B B B = Ω. We also know that P B =, P B =, and P B = 6. Finally we have, for the event A, that P A B i = i. We wish to find P B A, to do this we use Bayes rule, page 49. P B A = P A B P B P A B P B + P A B P B + P A B P B + = 6 8 = + 6 This is the rd option, X% chose this. Problem 7 Given the probability p =. of a plane being delayed from an airport, where it can be assumed that the delay of each plane is independent of the others. We wish to find the probability of at most 5 planes being delayed from an airport out of, choosing the random variable X as describing the number of delayed flights. To do this we use the normal approximation to the binomial distribution, page P X 5 Φ 4. This is the 5th option, X% answered this. Problem 8 Φ.86 Let X be a Uniform, distributed random variable. We define the random variable Y = log X. We find the density fy y by applying the formula on page 4, and setting

3 y = gx = log x. f Y x = d log X dx = x x = e y f Y y = e y This is the st option, X% answered this. Problem 9 Define the random variables X, Y, where X >, such that EY X = x = e x, and the marginal density of X is f X x = e x. We find EY by applying the formula on page 4. EY = EEY X = = This is the nd option, X% answered this. Problem EY X = xf X xdx e x dx = Let the random variables X,..X describing the size of each of the ten mink follow a normal9, distribution. We wish to find P X min 5. To do this we start by standardizing X, as shown on page 9, setting X = X 9, so we instead get P Xmin 5 9. This we can solve by using the formula for the distribution of the minimum of N random variables, page 9. P X = Φ To find this, we look up the value of Φ in Appendix 5, and use Φ z = Φz. We get Φ.6, which is the nd option. X% answered this. Problem We know that the probability of the event A i, choosing a drone in each draw, is p =, where i =..9. We wish to know the probability of having to look at at least ten bees before finding a drone, is found by application of the geometric distribution s tail probability, as summarized on page 48. Using this we find the probability of the event of not drawing any drones in the first ten tries P A c A c..a c 9 = 9. This is the rd option, X% chose this.

4 Problem Let A be the event that person has elevated bloodsugar, B the event that the patient has elevated blood pressure. We have P A = 5, P A B = 5, P A B = 4 5. We can find P B by using inclusion-exclusion, page. P A + P B P A B = P A B This is the rd option, X% answered this. Problem P B = P A B P A + P A B = = 5 Let p =.5 be the probability that each citizen arrives in the first half hour. Assuming the citizens arrivals to be independent of each other, we can describe the number that arrived by a random variable X which follows a binomial5,.5 distribution. We can thus find P X 4, by applying the relationship P X x = P X < x. Inserting into the binomial distribution, page 477, we get 5 P X 4 =.5 k.5 5 k k = 6 This is the 4th option, X% answered this. Problem 4 k= Let X, Y be random variables following a bivariate normal distribution, both with mean, tons and standard deviation of 5, tons, and correlation coefficient ρ = 5.We wish to find P X + Y,. We start by normalizing, page 9. X + Y,,, P = P X + Y 5 5 We then rewrite Y in terms of X and the independent standard normal variable Z, page 45, Y = 5 X + 9 5Z. Inserting this we get 8 P 5 X Z We then use the rule for addition of normally distributed random variables, page 6, and see that the above can be rewritten in terms of the random variable W norm, We can now finally find the probability, by normalizing with this new variance P W 5 8 = P W We then use the relationship P X x = P X < x This is exactly Φ, which is the 5th answer. 4

5 Problem 5 Let T be a random variable describing the lifetime of a component, we have: P t < T < t + t T > t lim = e t. t t The survival function P T > t can be found by application of the formula in the table on page 97, it is the 4th option. Problem 6 The probability of choosing exactly two surgeons who are knee specialists can best be described by a hypergeometric distribution, as this is exactly choosing g = good out of total pool N = containing G = 4 good. Problem 7 Given a circle centered in,, and considering the part of the circle that lies in the upper right quadrant, we wish to find the probability of a random point lying between the two lines y = x tan π 6, y = x tan π. Denoting the coordinates of the point X, Y, we find this probability by the method of areas as shown on page 4. The area bounded between the two lines is the arc of length θ = π 6, so by application of the formula for the area under an arc AC = θ r and inserting, we get AC = π. The area of the circle is π 4. We thus find, denoting the area between the lines AC, and the circle AD π AC AD = This is the 5th option, X% answered this. Problem π = The discrete random variable X follows a geometric distribution with parameter p, P X = x = p p x. The random variable Y is binomially distributed for a given X = x, with probability parameter q and number of trials x. We find P Y = by applying the formula for average condtional probability, page 44. P Y = = = p p x x!!x! p q x p p x q x n= n= = p q p x q x n= The sum is recognized as a geometric sum, as shown on page 56. This is the nd option, X% chose this. = p q p q 5

6 Problem 9 Let X,.., X 4 be exponentially distributed random variables with rate λ =. We wish to find the probability of the waiting time T = X + X + X + X 4, being T 5. This is recognized as exactly a gamma4, distribution, page page 86. Inserting into the right-tail probability formula on this page we find P T > 5 = k= This is the rd option, X% answered this. Problem exp 5 5k k! = 8 5 e Let X, Y follow a bivariate normal distribution, with EX = 6, EY = 55, SDX = SDY =.5, and correlation coefficient ρ =.85. We wish to find EY X = 6. To do this, we start by standardizing X, Y. X = X 6.5, Y = Y We then look up the the marginals of the bivariate normal distribution on page 45, and see that the distribution of Y given X is normalρx, ρ. Therefore the mean of Y for X =, ie. X=6, is.85, and therefore we have EY X = 6 = EY X = = This is the 5th option, X% answered this. Problem Let X be a binomial,.5 distributed random variable, and Y a Poisson distributed one, assumed independent. We wish to find EX 5 + Y 5. To do this we use the formula for a sum of expectations, page 67, and the expecations of a Poisson and binomial distribution on page 477. This is the st option, X% chose this. Problem EX 5 + Y = 5EX + EY = = 44.5 Let X, Y be independent random variables, describing respectively the number of air bubbles and grains pr. m. These occur with a frequency of pr. m, and the grains occur with a frequency of 5 pr. m independently of each other. This can be seen as an instance of the Poisson Scatter Theorem, page 8, as these can be seen as random hits with no overlap. Therefore we find that for an area of m, X P oisson, Y P oisson. We wish find the probability P X =, Y =, which we can separate as X, Y are independent, we can use the multiplication rule on page 5. This is the 4th option, X% answered this. P X =, Y = = P X = P Y = = e e = e 6

7 Problem We wish to find P X + Y = 7, where X, Y are random variables the largest and smallest of the results of two rolls with a fair 6-sided die. We realize that this is equivalent to just considering the two rolls, with the results denoted by Z, Z, so we need not think in terms of min-max. We know, that there are 6 possible combinations of Z, Z, and we wish to find P Z + Z = 7, so we just need to find the number of results giving 7. This can be done simply by listing all possible combinations giving 6, ie. writing the joint distribution table as shown on page 45. Z Z There are exactly 6, so P Z + Z = 7 = 6 6 = 6. Alternatively, it is initally noted that there can only be three cases, as the largest number must be either 4, 5 or 6. As the outcome space for all of these has two combinations, ie. the first roll of is one of these or the second one is, we must double the probabilities. We can now use the results of page 49. This gives us the following result: P X + Y = 7 = P X = 6, Y = + P X = 5, Y = + P X = 4, Y = = = 6 8 = 6 This is the rd option, X% answered this. Problem 4 Let X, Y be random variables with joint density fx, y = { e x+y for < x < y < x else. We wish to find P X, Y. To do this we integrate, as shown on page 49. P X, Y = x x = + e 6 4 e 4 This is equivalent to the second option, X% chose this. Problem 5 e x+y dydx A random variable X follows a beta, distribution with density f X x = 6x x. Further, we are given that for a discrete random variable Y, P Y = y X = x = y x y 7

8 x y. We determine P Y = by use of the integral conditioning formula, page 45, noting that the beta distribution is defined [; ]. P Y = = x x 6x xdx y = This is the rd option, X% answered this. Problem 6 We wish to find the probability that 5 draws with replacement from a set of 5 numbers are all different. We wish to find the probability of the event that all the 5 samples represent different categories. Let N i denote the event that the N i th sample is of a different category than the previous ones, the probability of the event N i, given that we have already drawn i different number is then given by 5 i P N i N N N...N i = P N i N i = 5 We can then find the desired probability, the probability that all five are not unique, which must be the complement to the problem that all five are different. P NNNN4N5 c = P N N N N 4 N 5 = This is the st option, X% answered this. Problem % = 4 5 i Let X, Y be standard bivariate normally distributed random variables, with ρ = 5. We wish to find P X < Y <. We start by rewriting Y in terms of the standard normal variable Z and X, as in the example on page 45. We find Y = 5 X Z, and insert this in the probability. P X < Y < = P X < 5 X + Z < 8 = P X < Z < 5 X This we find, also as in example, by finding the angle and dividing by the total of a circle. 8 P X < Z < 5 X = arctan arctan 5 π = arctan π This is the rd option, X% chose this. i= 5 8

9 Problem 8 Let R be a Rayleigh-distributed random variable describing the field strength at a given point. We wish to find P. < R <.. To find this we use the estimate, P X dx = fxdx, page 6. Setting x =, and dx =., inserting in the density on page 477, we find P < X <. e. = 5 e Alternatively, to find this we can use the CDF. of the Rayleigh distribution, page 478 P. < R <. = e. e. = e e. e 5 This is the 4th option, X% answered this. Problem 9 To determine whether two continous random variables are independent, one must check if their joint density can be written as the product of their densities, as shown on page 5. The marginal densities are determined from the joint densities. This is the nd option. X% chose this. Problem The random variables X, Y follow a bivariate distribution with joint density fx, y = { e x+y for < x < y < x ellers. We wish to find the density of a random variable Z = Y X, where X is unlimited. We find this by use of the approach on page 8, inserting directly in the formula as x is unlimited. We do however note that maxy x =, miny x =, so we have the bounds for < z <. fx, xz = e x+xz f Z z = = xe x+xz dx + z, < z < 9