Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
An exam or practice exam for a math 101 course, specifically covering module 3 content. It includes several probability-related questions, such as finding the probability of intersections and unions of events, calculating conditional probabilities, and solving combinatorics problems. Detailed step-by-step solutions to these problems, making it a valuable resource for students studying probability and statistics at the introductory university level. The content covers a range of topics, including basic probability concepts, bayes' theorem, and combinations, providing a comprehensive overview of the key concepts and techniques in this area of mathematics.
Typology: Exams
1 / 10
This page cannot be seen from the preview
Don't miss anything!
Suppose A and B are two events with probabilities: P(Ac^ )=.40,P(B)=.45,P(A∪B)=.60. Find the following: a) P(A∩B). p(anb) = p(a) + p(b) - p(aub) p(a) = 1 - p(a^c) p(a^c) = 0. 1 - 0.40 = 0. p(a) = 0. 0.60 + 0.45 - 0.60 = 0. P(AnB) = 0. b) P(A). p(a) = 1 - p(a^c) p(a^c) = 0. 1 - 0.40 = 0. P(A) = 0. c) P(Bc). p(b) = 1 - p(b^c) rearranged to find p(b^c) = 1 - p(b)
p(n,r) = n! / (n-r)! n = 8 r = 4 8 - 4 = 4 p(8,4) = 8! / 4! 8x7x6x5 = 1680 P(8,4) = 1680 1680 different passwords can be made out of the 8 characters given without one character showing up more than once
Suppose you are going to make a password that consists of 4 characters chosen from {2,7,8,c,f,k,t,z}. How many different passwords can you make if you cannot use any character more than once in each password?
Find the answer to each of the following by first reducing the fractions as much as possible: a) P(850,4)= b) C(530,4)=
In a tri-state conference, 60% attendees are from California, 25% from Oregon, and 15% from Washington. As it turns out 6 % of the attendees from California, 17% of the attendees from Oregon, and 12% of the attendees from Washington came to the conference by train. If an attendee is selected at random and found to have arrived by train, what is the probability that the person is from Washington? c = california o = oregon w = washington t = train p(c) = 60/100 = 0. p(o) = 25/100 = 0. p(w) = 15/100 = 0. p(t│c) = 0. p(t│o) = 0. p(t│w) = 0. p(t│w) = ((p(w) x p(t│w)) / ((p(w) x p(t│w)) + ((p(c) x p(t│c)) + ((p(o) x p(t│o)) (0.15 x 0.12) / ((0.15 x 0.12) + (0.60 x 0.06) + (0.25 x 0.17)) 0.018 / (0.018 + 0.036 +0.0425) 0.018 / 0.0965 = 0. b) What is (B│A)?
Exam Page 6 The probability that a certain type of battery in a smoke alarm will last 2 years or more is .85. The probability that a battery will last 5 years or more is .15. Suppose that the battery is 2 years old and is still working, what is the probability that the battery will last at least 5 years? e = battery lasting 2 or more years f = battery lasting 5 or more years p(e) = 0.
In a tri-state conference, 60% attendees are from California, 25% from Oregon, and 15% from Washington. As it turns out 6 % of the attendees from California, 17% of the attendees from Oregon, and 12% of the attendees from Washington came to the conference by train. If an attendee is selected at random and found to have arrived by train, what is the probability that the person is from Washington? P(Train│C)=.06.. P(Train│O)=.17.. P(Train│W)=.12.. P(C)=.60,P(O)=.25,P(W)=.15. We want to find P(W│Train), so use:
combo order not important c(n,r) = n! / r!(n-r)! n = r = 5 11 - 5 = 6 c(11,5) = 11! / (5!)(6!) (11x10x9x8x7) / (5x4x3x2x1) 55440 / 120 = 462 C(11,5) = 462 when order is not important, there are 462 ways 5 out of 11 people can be chosen to go on a mission trip. Answer Key Suppose that 5 out of 11 people are to be chosen to go on a mission trip. In how many ways can these 5 be chosen if the order in which they are chosen is not important. Since we do not want to count all of the possible orderings, we use combinations.
