M3 Problem Set

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M3: Problem Set Due No due date Points 5 Questions 11 Time Limit None Attempt History Attempt Time Score LATEST Attempt 1 149 minutes 5 out of 5 Score for this quiz: 5 out of 5 Submitted Dec 3 at 7:04pm This attempt took 149 minutes. Question 1 0/0pts According to the National Oceanic and Atmospheric Administration (NOAA), between 1851 and 2013 there were 290 hurricanes that hit the U.S. Coast. Of these, 117 were Category 1 hurricanes, 76 were Category 2 hurricanes, 76 were Category 3 hurricanes, 18 were Category 4 hurricanes, and 3 were Category 5 hurricanes. Make a probability distribution for this data. If a hurricane hits the U.S. coast, what is the probability that the hurricane will be a Category 1 hurricane. Your Answer: Category|Probability 1 117/290 = 0.403 76/290 = 0.262 76/290 = 0.262 18/290 = 0.062 5/290= 0.011 o~ ODN There is a .403 probability that it will be a category 1 hurricane.

Solution: Category Probability 117/290 = 403 76/290 = 262 76/290 = 262 18/290 = .062 3/290 = 011 Q| B QO N — We can see from the probability distribution, that if a hurricane hits the U.S. coast, there is a .403 probability that it is a category 1 hurricane. Question 2 0/0 pts Find the answer to each of the following by first reducing the fractions as much as possible: a) P(17,8)= b) C(19,15)= Your Answer: a) P (n,r)=n!/ (n-r)! P(17, 18) = 171/ (17-8)! = 171/9! = 17(16)(15)(14)(13)(12)(11)(10) = 9 b) C(n, r) = n!/ rl(n-r)! C(19, 15) = 19! / 151(19-15) = 19! / 15141 = 19(18)(17)(916) / 4(3)(2)(1) = 3876

Solution. a) For permutations, we will use: n! P(n, r) - m o oSy 5 = 980,179,200 P(17,8) = ar—e)l_ o 17 (16)(15)(14)(13)(12)(11)(10) = ,179, b) For combinations, we will use: n! i i ey B 19! _ 190 19 (18)(17)(16) _ €(19.15) = 15! (19 —15)! 15!4! 4(3)(2)(1) 3876 Question 3 0/0 pts Suppose you are going to make a password that consists of 5 characters chosen from {2,3,6,7,a,c,k,n,p}. How many different passwords can you make if you cannot use any character more than once in each password? Your Answer: Ordering matters P(n, r) = N!/ (n-r)! 91/ (9 - 5)! = 9! / 41 = 9(8)(7)(6)(5) = 15,120

Solution. For passwords, order matters. For example, the password 63pn2 is different than the password p2n36 even though both passwords use the exact same characters. Therefore, we want to count the total number of ways that we can choose and order the characters. That means that we should use permutations. We have a total number of nine characters to choose from, so n=9. We will be selecting five characters, so r=5. We will use: n! P(n,r) = =) ! 9! P(9,5) = @35 = Fm = 9(8)(7)(6)(5) = 15,120. Question 4 0/0 pts Suppose that 10 countries submit bids for the summer Olympics. The Olympic committee will select 4 finalists out of these 10 countries. In how many ways can the Olympic committee pick these 10 finalists? Your Answer: Order does not matter C(n, r) = n!'/ ri(n-r)! 10! / 41(10-4)! = 10! / 416! = 10(9)(8)(7) / 4(3)(2)(1) = 210

Solution. Here order does not matter. So, we will use combinations. There are ten countries to choose from, so n=10. Four countries will be chosen, so r=4. We will use Gl rl (n - r)! 10t 10t 10(9)(8)(7) €104) = 41 (10—-4)! 416! 4(3)(2)(1) = Question 5 0/0 pts Suppose A and B are two events with probabilities: P(A) =.40,P(B) =.45,P(AN B) =.25. Find the following: a) P(A U B). b) P(A°). c) P(B°). Your Answer: P(A) = .40, P(B) = .45, P(B) - 25 a) PAlU[B) = P(a) + P(B) - P(AINB) 40+.45-.25=.6 b) P(A)=1-P(Ac) converted to (Ac)=1-P(A) 1-.40=.60

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