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A series of exercises and solutions related to probability and combinations in mathematics. It covers topics such as calculating probabilities of events, understanding conditional probability, and applying the concepts of permutations and combinations to solve problems. Step-by-step solutions for each exercise, making it a valuable resource for students studying probability and combinatorics.
Typology: Exams
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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) 1 - 0.45 = 0.
P(B^c) = 0.
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). For P(A∩B). Use P(A∪B)=P(A)+P(B)-P(A∩B) and rearrange to P(A∩B)=P(A)+P(B)-P(A∪B). But for this equation, we need P(A) which we can find by using P(A)=1- P(A^c ). So, P(A)=1-.40= .60. P(A∩B)=.60+.45-.60=.45. b) P(A). P(A) was found above as .60. c) P(Bc). For P(Bc^ ). Use P(B)=1-P(Bc^ ) which may be rearranged to (Bc^ )=1-P(B). P(Bc^ )=1-.45=.55.
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)=
Suppose A and B are two events with probabilities: P(Ac^ )=.10,P(B)=.75, P(A∩B)=.50. a) What is (A│B)? p(a│b) = p(anb) / p(b) p(anb) = 0. p(b) = 0. 0.50/0.75 = 0. P(A│B) = 0. b) What is (B│A)? p(b│a) = p(anb) / p(a) p(anb) = 0. p(a) = 1 - p(a^c) p(a^c) = 0. 1 - 0.10 = 0. p(a) = 0. 0.50 / 0.90 = 0. P(B│A) = 0.
Suppose A and B are two events with probabilities: P(Ac^ )=.10,P(B)=.25,P(A∩B)=.50. a) What is (A│B)?
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│W)=.12.. P(C)=.60,P(O)=.25,P(W)=.15. We want to find P(W│Train), so use: 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.
p(f) = 0. p(enf) = battery greater than 5 years = 0. p(f│e) = p(enf) / p(e) 0.15/ 0.85 = 0. P(F│E) = 0. Answer Key The probability that a certain type of battery in a smoke alarm will last 2 years or more is .85. The probability that the 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? Define E to be the event that the battery will last 2 years or more. Define F to be the event that the battery will last 5 years or more. We are given that: P(E)=.85. P(F)=.15. Note also that if a battery lasts more than 5 years, it would have had to have lasted 2 years, so P(E∩F)=.15. The question is asking “given event E has occurred, what is the probability that event F will occur”? This may be found by using: Exam Page 7 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.
