MATH 446/546 Homework 1:

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1 MATH 446/546 Homework 1: Due September 28th, 216 Please answer the following questions. Students should type there work. 1. At time t, a company has I units of inventory in stock. Customers demand the product at a constant rate of d units per year (assume that I d). The cost of holding 1 unit of stock in inventory for time is $h. Determine the total holding cost incurred during the next year. The inventory at time t is I td. The holding cost will then be from time t to time t + can be represented as: Holding Cost h(i td) Thus the total holding cost for the next year can be considered as: Holding Cost Year h 1 1 h(i td) dt hi hdt dt ( hit hd 2 t2 ( I d ) 2 ) 1 The holding cost for the next year is given by h ( I d 2) dollars. 2. What is the probability that the total of two dice will add up to 7 or 11? Consider the possible out comes for rolling a single die: {1, 2, 3, 4, 5, 6} This gives 36 possible out comes corresponding to a value given for the first die paired with the different possibilities that can be obtained on the second die. Considering the ways the die can add to 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

2 with the ways that the dice can add to 11 gives: (5, 6), (6, 5) or 8 of the possible 36 outcomes for the different die rolled. Thus the probability of rolling a 7 or an 11 is Three out of every 1, low-risk 5 year old males have colon cancer. If a man has colon cancer, a test for hidden blood in the stool will indicate hidden blood half the time. If he does not have colon cancer, a test for hidden blood in the stool will indicate hidden blood 3% of the time. If the hidden-blood test turns out positive for a low-risk 5-year old male, what is the chance that he has colon cancer? This requires the use of Bayes Rule. Define We aim to compute: P (CC) 3 : Probability of Colon Cancer P (ĈC) : Probability of no Cancer 1 P (T CC).5 : Testing positive given Colon Cancer P (T ĈC).3 : Testing positive given No Colon Cancer Here we need Bayes Rule: P (CC T ) P (CC T ) P (T CC)P (CC) P (T CC)P (CC) + P (T ĈC)P (ĈC) P (CC T ).5(3/1).5(3./1) +.3(997/1) If the hidden-blood test turns out positive for a low-risk 5-year old male the chance that he has colon cancer is approximately An urn contains 1 red balls and 3 blue balls. (a) Suppose you draw 4 balls from the urn. Let X i be the number of red balls on the i th ball (X i or 1). After each ball is drawn, it is put back into the urn. Are the random variables X 1, X 2, X 3, and X 4 independent random variables? Page 2

3 (b) Repeat part (a) for the case in which the balls are not put back in the urn after being drawn. For part (a) not that the balls are replaced after each draw thus, the probability of drawing a red ball P (X i 1) 1 and P (X 4 i ) 3 for all draws i, then we 4 know the X is are independent. For part (b) we make note that we need: P (X 1 1 X 2 1) P (X 1 1)P (X 2 1) in order for the events to be considered independent. Here we have: P (X 1 1) 1/4, and P (X 2 1) (9/39)(1/4) + (1/39)(3/4).25 Thus, and P (X 1 1)P (X 2 1).625 P (X 1 1 X 2 1) (1/4) (9/39) Weekly Ford sales follow a normal distribution with a mean of 5. cars and a standard deviation of 14. cars. (a) Find X if there is to be a 1% chance that Ford will sell more than, X cars during the next year. (b) What is the chance that Ford will sell between 2.4 and 2.7 million cars during the next year? We need to consider the translation of the weekly normal distribution to an annual normal distribution. Note that the weeks do not overlap and hence we can consider the random variable X i as the sales of Ford during week i of the year and is distribute as X i N(µ i 5., σ 2 i wvar) Then the annual sales is distributed as: ( X year N µ year µ i, σyear 2 i1 i1 σ 2 i ) Page 3

4 X year N ( µ year 26., σ 2 year ) We now need to find the value of X that corresponds to a lower tail of.99 with the yearly mean and standard deviation just found. Here we want the X corresponding to a lower tail of size.99. Thus, Z X Zσ + µ Thus the number of cars to sell so that there is only a 1.% chance that ford will sell more than that number of cars is The plot shows the area shaded on a standard normal plot representing part (a) for the problem. For the second part of the problem we need to consider Z scores for 2.4 and 2.7 million cars: Z µ σ Z µ σ Integrate the standard normal curve between and.995 gives Thus the probability that Ford will sell between 2.4 and 2.7 million cars during the next year is A graphical visualization of this is: Some t years form now, air conditioners are sold at a rate of t per year. How many air Page 4

5 conditioners will be sold during the next five years? Here we integrate the rate function to get the total number of air conditioners sold: T otal 5 t dt 1 2 t We expect to sell 12.5 air conditioners in the next 5 years. 7. When three friends go for coffee, they decide who will pay the check by each flipping a coin and then letting the odd person pay. If all three flips are the same (so there is no odd person) then they make a second round of flips, and continue until there is no odd person. What is the probability that: (a) Exactly 3 rounds of flips are made? (b) More than 4 rounds of flips are needed? (a) Here we consider the different possible out comes of the flipping: {HHH, HHT, HT H, T HH, HT T, T T H, T HT, T T T } Note that the probability of an odd person on the flip is P (O) 3/4, and the probability of all coins coming up the same is P (S) 1/4. Note each round of flipping is independent of the round before. P (3round) P (S)P (S)P (O) (1/4) 2 (3/4) The probability of exactly three rounds of flipping is.47. (b) To find the probability of more than 4 rounds being needed consider the complement and subtract from 1. Thus, P (More Than 4 rounds) 1 P (exactly one round) P (exactly two rounds) Finding the probability of exactly two rounds is: P (exactly three rounds P (exactly four rounds) P (exactly two rounds) (1/4)(3/4) Page 5

6 Finding the probability of exactly four rounds is: Thus, P (exactly two rounds) (1/4) 3 (3/4) P (More Than 4 rounds) 1 (3/4) Additional Homework for 546 Students 1. Suppose that there are n possible outcomes of a trial, with outcome i resulting with probability p i, i 1,..., n, n i1 p i 1. If two independent trials are observed, what is the probability that the result of the second trial is larger than that of the first? Let L 1 and L 2 be the events that the first trial and the second trial is larger respectively. Then we can let E be the event that the two probabilities are equal. Then note as these events are exclusive: P (L 1 ) + P (L 2 ) + P (E) 1 Also note that there probability of L 1 is equal to the probability of L 2 by symmetry. Then P (L 2 ) 1 P (E) 2 1 n i1 p2 i 2 A second way to think about the problem is to note that P (L 2 ) P ( The first gives i, the second trial gives j) i j>i i j>i Page 6

7 These values can easily be shown to be the same: 1 i1 i1 p i j1 p j j1 p 2 i + i1 p 2 i + 2 i1 j i i1 i1 j>i 2. There are two local factories that produce radios. Each radio produced at factory A is defective with probability of.5, whereas each one produced at factory B is defective with probability.1. Suppose you purchase two radios that were produced at the same factory, which is equally likely to have been either factory A or factory B. If the first radio that you check is defective, what is the conditional probability that the other one is also defective? Let D i be the event that radio i is defective. Let A and B be the events that the radios were produced at factory A and B respectively. Then, P (D 2 D 1 ) P (D 1D 2 ) P (D 1 ) P (D 1D 2 A)P (A) + P (D 1 D 2 B)P (B) P (D 1 A)P (A) + P (D 1 B)P (B).52 (1/2) (1/2).5(1/2) +.1(1/2) 13/3 Page 7