# MATH 201. Final ANSWERS August 12, 2016

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1 MATH 01 Final ANSWERS August 1, 016 Part A points) A bag contains three different types of dice: four 6-sided dice, five 8-sided dice, and six 0-sided dice. A die is drawn from the bag and then rolled. Assume that each die is equally likely to be drawn from the bag. a) What is the probability of rolling a? b) If the outcome of the roll is a 5, what is the probability that an 8-sided die was rolled? a) Let X denote the outcome of the roll. Let E 1 be the event of rolling a six-sided die, E the event of rolling an 8-sided die, and E 3 the event of rolling a 0-sided die. We are given that P E 1 ) = 4/15, P E ) = 5/15 and P E 3 ) = 6/15. By writing Ω = E 1 E E 3, we find P X = ) = P X = E 1 )P E 1 ) + P X = E )P E ) + P X = E 3 )P E 3 ) = 04/15) + 05/15) + 1/0)6/15) = 1/50. b) Using Bayes formula, we find P X = 5 E )P E ) P E X = 5) = P X = 5 E 1 )P E 1 ) + P X = 5 E )P E ) + P X = 5 E 3 )P E 3 ) 1/8)5/15) = 1/6)4/15) + 1/8)5/15) + 1/0)6/15) = points) Suppose X is a random variable such that EX = 4 and VarX = 3. Compute E3X ) ). 1

2 We compute E3X ) ) = E9X 1X + 4) = 9EX 1EX + 4 = 9VarX) + EX) ) 1EX + 4 = ) 14) + 4 = points) The floor function x = max{n Z : n x} gives the largest integer less than or equal to x. For example 3.5 = 3 and 4 = 4. Suppose X is a continuous random variable with probability density function given by 1 if x 1, x fx) = 0 if x < 1. a) Find the probability mass function p X k) for X. b) Is E X <? Justify your answer. a) Since X 1 with probability 1, we clearly have p X k) = P X = k) = 0 for all integer k 0. For each integer k 1 we find p X k) = P X = k) = P X [k, k + 1)) = = 1 kk + 1). k+1 k [ dx x = 1 ] k+1 = 1 x k k 1 k + 1 b) No. Using part a) we compute E X = kp X = k) = k=1 1 k kk + 1) = 1 k + 1, k=1 k=1 and this is the divergent harmonic series, indicating that E X = points) Suppose you play cards with a standard 5-card deck. In this game, 4 players are each dealt 13 cards, with all outcomes equally likely. A standard deck of cards contains 4 suits: clubs, diamonds, hearts, and spades. Each suit has the cards through 10, and a jack, queen, king, and an ace.

3 a) What is the chance that you get all 4 aces and all 4 kings? b) What is the chance of getting aces and kings? a) It seems best to set up the probability space using sampling without replacement, order doesn t matter. There are 5 choose 13 ways to get a 13-card hand. There is only one way to get 4 aces and 4 kings, and 5-8) choose 5 ways of getting the 5 remaining cards. Therefore, the probability is 44 ) 5 ) 5 13 b) We set up the probability space the same way as in part a). There are 4 choose ways of getting the aces, and 4 choose ways of getting the kings. There are 5-4) choose 9 ways of getting the remaining 9 cards. So the probability is 4 ) 44 ) 9 5 ) points) Consider the circuit drawn below. 1 A 4 3 B Electricity flows from A to B if there is at least one path available. Let p i be the probability that switch i is closed, so electricity can flow through it. Assume that the switches are open or closed independently of each other. In terms of p 1,..., p 4, what is the probability that electricity can flow from A to B? Hint: You may want to use the inclusion-exclusion formula. Let A i be the event that switch i is closed. Let F be the event that electricity can flow 3

4 through the circuit. Since there are 3 possible paths for the electricity, we have F = A 1 A ) A 1 A 3 ) A 4 Using the inclusion-exclusion formula and the independence of switches, we get P F ) =P A 1 A ) + P A 1 A 3 ) + P A 4 ) P A 1 A A 3 ) P A 1 A A 4 ) P A 1 A 3 A 4 ) + P A 1 A A 3 A 4 ) =p 1 p + p 1 p 3 + p 4 p 1 p p 3 p 1 p p 4 p 1 p 3 p 4 + p 1 p p 3 p points) Suppose that you are snowed in on a Saturday night at the dorms, and to kill time you flip coins. Usually they land on heads or tails, but on rare occasions they land on their edge. The probability of getting an edge on a given flip is 1/100, and these events are independent. Let A be the event that you get 55 edges in 5000 flips. a) Write an expression for the exact probability of A. b) Use the Poisson approximation to approximate the probability of A. a) Using the binomial probabilities, 5000 P A) = 55 ) ) 55 ) b) Using the Poisson approximation for the binomial, we find that λ = np = 5000/100 = 50, so Part B P A) ! e points) Suppose that the joint probability mass function p X,Y i, j) = pi, j) for random variables X, Y is given as follows. p1, 1) = 0.1 p1, ) = 0. p1, 3) = 0.3 p, 1) = 0.1 p, ) = 0. p, 3) = 0.1 a) Find EX. 4

5 b) Find EY. c) Find CovX, Y ) a) According to the above table, X only takes on the values 1 and. Summing the probabilities in the first row of the table, we find that P X = 1) = 0.6. Summing the second row, we find P X = ) = 0.4. So EX = = 1.4 b) Summing the probabilities in the columns of the table, we find that P Y = 1) = 0., P Y = ) = 0.4, P Y = 3) = 0.4. Thus, EY = =. c) Recall that CovX, Y ) = E[XY ] EX)EY ). First we compute E[XY ] using the table. E[XY ] = = 3 Therefore CovX, Y ) = E[XY ] EX)EY ) = = points) A store opens at 8:00am. Assume that customers arrive according to a Poisson process with parameter λ, with time measured in minutes. Suppose that the last customer to arrive before 9:00am arrives at T minutes before 9:00am. If no customers arrived between 8:00am and 9:00am, let T = 60 1 hour). Find the cumulative distribution function cdf) of T. Hint: Recall that P T > t) = 1 F T t). First, it is clear that T 0 with probability 1, so F T t) = 0 if t < 0. So assume that t 0. Also, when t > 60, since T is at most 60, we get F T t) = P T t) = 1. Since customer arrivals are modeled by a Poisson process NI), we see that if 0 t 60 then P T > t) = P N[60 t, 60]) = 0) = e λt. 5

6 So, if 0 t 60 then F T t) = 1 P T > t) = 1 e λt. To summarize, 0 if t < 0 F T t) = 1 e λt if 0 t 60 1 if t > points) Suppose that X 1, X,... are i.i.d. independent and identically distributed) with mean µ = 3 and variance σ = 5. Recall that X = X 1+ +X n n µ. For n = 95, approximate P X 3.5) is a common estimator for Express your answer in terms of Φx), the cumulative distribution function of the standard normal distribution for positive values of x. We let S n = X X n, and recall that the central limit theorem says that for large values of n, S n nµ σ n is approximately distributed according to the standard normal distribution. Therefore, P X 3.5) = P S ) n n 3.5 We subtract µ = 3 from all 3 terms on the right, divide by σ = 5, and finally multiply by n = 95. This gives us P S ) n n 3.5 = P 3 S n nµ n 3 = P = P 5 S n nµ σn ) ) 5 3) 95 S n nµ 5 σ n = P S ) n nµ σ n P ) Z ) 3.5 3)

7 where Z is a N0, 1) random variable. Expressing the probability in terms of Φ, we get P S ) n n 3.5 P ) Z ) = Φ Φ ) ) [ )] = Φ 1 Φ ) ) = Φ + Φ points) Consider the triangle D with vertices 0, 0), 1, 0) and 1, 1). Suppose X, Y ) is a random point chosen uniformly from the triangle D. a) Find the marginal densities of X and Y. b) Are X and Y independent? Justify your answer. y 1, 1) D 0, 0) 1, 0) x a) Since the triangle has area 1/, the joint probability density function of X, Y ) is given by if x, y) D, f X,Y x, y) = 0 if x, y) / D. 7

8 We compute the marginal densities through integration x dy if x [0, 1] 0 f X x) = f X,Y x, y)dy = 0 if x / [0, 1] x if x [0, 1] = 0 if x / [0, 1]. 1 dy if y [0, 1] y f Y y) = f X,Y x, y)dx = 0 if y / [0, 1] 1 y) if y [0, 1] = 0 if y / [0, 1]. b) Note that f X,Y x, y) f X x)f Y y) for some x, y) D, and it follows that X and Y are not independent points) Suppose that X and Y are independent random variables. Assume that X has range {1,, 3} with probability mass function P X = 1) = 1 4 P X = ) = 1 P X = 3) = 1 4, and assume that Y has range {, 3, 4} with probability mass function P Y = ) = 1 P Y = 3) = 1 6 P Y = 4) = 1 3. a) Find the moment generating functions M X t) and M Y t) of X and Y, respectively. b) Compute M X+Y t) and use this to identify the range and probability mass function of X + Y. a) We compute M X t) = Ee tx ) = e t P X = 1) + e t P X = ) + e 3t P X = 3) = 1 4 et + 1 et e3t, 8

9 and M Y t) = Ee ty ) = e t P Y = ) + e 3t P Y = 3) + e 4t P Y = 4) = 1 et e3t e4t. b) Since X and Y are independent, we find 1 M X+Y t) = M X t)m Y t) = 4 et + 1 et + 1 ) 1 4 e3t et e3t + 1 ) 3 e4t = 1 8 e3t e4t e5t e6t e7t. It follows that X + Y has range {3, 4, 5, 6, 7} with the probability mass function P X + Y = 3) = 1 8 P X + Y = 6) = 5 4 P X + Y = 4) = P X + Y = 5) = 7 4 P X + Y = 7) = points) Suppose X Poisson6) and Y Exp1/6) with CovX, Y ) = 1. a) Use Markov s inequality to obtain an upper bound on P X + Y > 0). b) Find the variance VarX + Y ). c) Use Chebyshev s inequality to obtain another upper bound on P X + Y > 0). a) Recall that if X Poissonλ), then EX = λ and VarX = λ. In this case EX = VarX = 6. Also, if Y Expλ), then EX = 1/λ and VarX = 1/λ. In this case EY = 6 and VarY = 36. Note that X 0 and Y 0, so we also have X + Y 0. By Markov s inequality, we find P X + Y > 0) EX + Y ) 0 = EX + EY 0 = 1 0 = 3 5. b) We compute VarX + Y ) = CovX + Y, X + Y ) = VarX) + CovX, Y ) + VarY ) = 6 + 1) + 36 = 18. 9

10 c) By Chebyshev s inequality, since EX + Y ) = 1, we find P X + Y > 0) = P X + Y > 1 + 8) P X + Y 1 > 8) VarX + Y ) 8 = =

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