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A series of probability exercises and problems designed to reinforce understanding of key concepts in probability theory. It covers various aspects of probability, including calculating probabilities of events, understanding conditional probability, and determining independence of events. The exercises are presented in a clear and concise manner, making them suitable for students learning about probability for the first time or for those seeking to solidify their understanding.
Typology: Assignments
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a. The P(Adam loses) = 1 – P(Adam wins) = 1 – 0.42 = 0. b. 0.31 the answer is because the P(either Brown or Dalton wins = P(Brown wins) + P(Dalton wins) = 0.09 + 0.22 = 0. c. P(either Adams, Brown or Collins wins = P(Adamas wins) + (Brown wins) + P(Collins wins) = 0.
a. The sample space for the number of computer sold is, S = { 0,1,2,3,4,5} b. An event E sells more than three computers is E = {4,5} A. The rate of promotion among female assistant professor is, (0.03) / (0.03+0.12) * 100% = (0.03)/(0.15) * 100% = 20% B. The rate of promotion among male assistant professor is, (0.17)/(0.17+0.68) * 100% = (0.17)/(0.85) * 100% = 20% C. By overserving the rate of promotions among female and male assistant professors, it can be concluded that no gender biasedness in the university We are omitting Questions A and B
A. From the given joint probability distribution table, the proportion of customers say that they will return and rate the restaurant’s food as good is 0. B. (0.35) / (0.02+0.08+0.035+0.2) = 0.54. Therefore, the proportion of customers who say that they will return rate the restaurant’s food as good is 0. C. (0.35) / (0.35+0.14) = 0.71 Therefore, the proportion of customers who rate the restaurant’s food as good say that they will return is 0. D. From the above calculations, it is observed that for part (a), we use joint probability value to computer the required value and for parts (b) and (c) we use conditional probability to compute the required proportion values.
B. P(B1 and A2) = 0. C. (0.22) / (0.50) = 0. D. (0.27) / (0.49) = 0.55 And P(A1)= 0.5 Since the two probabilities are not equal, then we can conclude that the two events are not independent.
a. What is the probability that a home is listed for more than 90 days before being sold? 200 / 800 = 0. b. What is the the probability of a home in this large metropolitian area pricing under $150,000? 100 / 800 = 0. c. What is the probability a home is listed for more than 90 days and pricing is under $150,000? 10 / 800 = 0.
