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This Homework Will Not Be Collected Nor Graded NAME (Please Print): KEY Statistics 110 Homework 3 You are allowed to discuss problems with other students, but the final answers must be your own work. For all problems that require calculation, YOU MUST ATTACH SEPARATE PAGES, NEATLY WRITTEN, THAT SHOW YOUR WORK. Please mark your answer in the space provided. As a general rule, each blank counts for one point. If necessary work is not shown, or if that work is substantially wrong, then you will not get credit even if the answer is correct. (The obvious purpose of this seemingly draconian policy is to prevent people from mindlessly copying each other’s answers.) Report all numerical answers to at least two correct decimal places. THIS HOMEWORK WILL NOT BE COLLECTED NOR GRADED. 1 Let A = 1, 2, 3, depending on whether the first letter of your last name is in A-F, G-S, or T-Z, respectively. 1. A standard poker deck has 52 cards, in four suits (clubs, diamonds, hears, spades) of thirteen cards each (2, 3, ..., 10, Jack, Queen, King, Ace, in ascending order). .0020 What is the probability of being dealt a flush (all five cards of the same suit)? Give two accurate non-zero decimal places. Note: Consider a straight flush to be a flush. For a particular suit, this is (13/52) * (12/51) * ... *(9/48). Since there are four possible suits, multiply this by 4 to get .001980792. .0039 What is the probability of being dealt a straight (all five cards are in consecutive order)? Give two accurate non-zero decimal places. (Note: An ace may be low or high in a straight, but not both, which precludes straights that “wrap-around”.) Note: Consider a straight flush to be a straight. Once we pick a beginning value, the probability of getting a straight that starts at that value and runs consecutively is (4/52) * (4/51) * ... * (4/48). But the cards don’t have to arrive in consecutive order—there are 5! = 120 ways they could be dealt. And there are 10 possible beginning values: Ace, 2, 3, ..., 10. Thus the answer is 10 * 120 * (4/52) * ... * (4/48) = .00394. .0059 What is the probability of being dealt a straight or a flush? Give two accurate non-zero decimal places. The probability of A or B is P[A] + P[B] - P[A and B]. We have the first two terms from the previous answers. The probabilty of A and B is the probabilty of a straight flush: 10 starting points * 5! orderings * (4/52) * (1/51) * ... * (1/48). The last bit arrives because any of the four ssuits are possible for the first card that is dealt, but after that each of the subsequent four cards has to be of that suit. This is .00001539. So subtracting this from the sum of the two previous answers gives .005905. You are dealt the king of hearts, the ten of spades, the eight of hearts, the three of spades, and the jack of diamonds. You discard the ten, the eight, and the three. What is the chance that your final hand has exactly A kings (where A is the number corresponding to your last name)? Give two accurate non-zero decimal places. 2 P[ 1 king ] = probability that the next three draws are not kings, or (44/47) * (43/46) * (42/45) = .81677. P[2 kings] = probability that exactly one of the next three draws is a king, or (3/47) * (44/46) * (43/45) * 3 = 0.175. The last multiplication by 3 reflects the fact that the king could be the first, second, or third card drawn. If this does not make sense, add up the probabilities you would get from the three incompatible events: getting the king first, getting the king second, and getting the king third. It will add up to what we got. P[3 kings ] = probability that exactly two of the next three draws are kings, or (3/47) * (2/46) * (44/45) * 3 = 0.00814, by similar logic to the foregoing. 2. Abelard tutors Heloise, the niece of Canon Fulbert, in theology. But they fall in love. Suppose they kiss each time they encounter the word “tropological” in their text. What would be a good probability model for the number of times they kiss in an hour-long study session? Which model? Poisson Why? No limit on the number; fixed amount of time. (But probably not independent, since a topic in a text might appear in bursts, so this is a bit of an approxima- tion...) Suppose that on average, they kiss 4.5 times. What is the probability that they kiss A + 3 or more times tomorrow? Probability=1-P(Kiss<A+3). And 0 1 A+2 4.5 − . 4.5 − . 4.5 − . P ( Number of Kisses < A +3) = e 4 5 + e 4 5 + . + e 4 5 0! 1! (A + 2)! For A = 1, probability is 0.658. For A=2, prob=0.4679, For A=3, prob=0.297. Canon Fulbert discovers the romance and hires A + 3 bravos to catch Abelard and castrate him. It takes two bravos to apprehend him; if just one grabs him, then Abelard can knock him down and escape. And suppose Abelard is faster than A/(A + 1) ∗ 100% of the population. 3 What probability model do you use? binomial What is the probability that Abelard escapes? A A+3 A+3 A A+2 P(k=0 or 1)=( A+1 ) + A+1 ( A+1 ) For A=1, Prob=5/16=0.3125. For A=2, Prob=0.4609. For A = 3, prob=0.5339 3. Suppose you are inspecting a shipment of 50 lightbulbs. You do not have time to test them all, so your protocol is to select A + 3 at random, and reject the shipment if more than one is defective. If 10 of the bulbs are defective, what is the probability that you reject the shipment? Let X be the number of lightbulbs that fail. We want to find IP(X = 0) + IP(X = 1), the probability that the shipment is accepted, and subtract this from 1. The number that fail follows a hypergeometric distribution. For A = 1. 10 40 0 4 IP[X =0]= = 0.3968 50 4 and 10 40 1 3 IP[X =1]= = 0.4290 50 4 and so the probability of rejecting the shipment is 0.1742. 4. Suppose a lightbulb has probability A/(A + 2) of working. You need 3 working lightbulbs for a science fair project. What is the probability that you must test more than 5 lightbulbs? This is a negative binomial distribution. You want to find the probability that X > 2, where X is the number of failures before the third lightbulb is obtained. The formula is x + r − 1 r x IP[X = x]= p (1 − p) r − 1 where r = 3. For A = 1, then p = 1/3 and thus 2 3 0 IP[X =0]= (1/3) (2/3) = 0.0370 2 4 and 3 3 1 IP[X =1]= (1/3) (2/3) = 0.0741 2 and 4 3 2 IP[X =2]= (1/3) (2/3) = 0.0988. 2 Thus the probability that X > 2 = 1 - (0.0370) - (0.0741) - (0.0988)= 0.79. For A = 2, the answer is 0.5; for A = 3 it is 0.32. 5. Pittsburgh Plate Glass, Inc., claims that the number of defects in a square yard of plate glass is, on average, less than 0.5. Before doing business with them, you decide to inspect 2 square yards of their glass. If you find more than A + 1 defects, you will not sign a contract. Suppose PP&G has exactly 0.5 defects per square yard on average. What is the probability that you do not sign the contract? (In hypothesis testing, this is called the Type I error rate, or the α level of your test.) Since you are looking at 2 square yards, the defect rate is λ = 1. The number of defects is Poisson. So the probability of not signing the contract is 1 − IP[X ≤ A+1 1x A+1] = 1− x=0 x! exp −1. For A = 1, 2, 3 this is 1−0.919, 1−0.981, 1−0.996, or 0.081, 0.019,P and 0.0037. Suppose PP&G has exactly 1.5 defects per square yard on average. What is the probability that you do not sign the contract? (In hypothesis testing, this is called the power of your test.) A+1 3x If λ = 2 ∗ 1.5, then IP[X > A + 1] = 1 − IP[X ≤ A + 1] = 1 − x=0 x! exp −3. For A = 1, 2, 3 this gives 1-0.423=0.577, 1-0.647=0.353, and 1-0.815=0.P 185, re- spectively. You inspect 2 square yards and find A + 1 defects. What is the probability of finding this many defects or more if PP&G’s claim is correct? (In hypothesis testing, this is called the significance probability.) This is like the first part, except now you sum from A + 1 and up, instead of A 1x fromA +2 and up. 1 − IP[X ≤ A] = 1 − x=0 x! exp −1. For A = 1, 2, 3 this is 0.264, 0.080, and 0.019, respectively. P 5 6. Cash-strapped Durham has asked you to design a lottery to bring in money. It costs $1 to play. Players pick 5 distinct numbers from the integers 1 to 25 inclusive.
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