2.14 - (c) Let F denote fatal accident. We have 50% of (30...

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2.14 (a) Note that the question assumes an accident either occurs or does not occur at a given crossing each year, i.e. no more than one accident per year. There were a total of 30 + 20 + 60 + 20 = 130 accidents in 10 years, hence the yearly average is 130 / 10 = 13 accidents among 1000 crossings, hence the yearly probability of accident occurring at a given crossing is 1000 13 = 0.013 (probability per year) (b) Examining the data across the “Day” row, we see that the relative likelihood of R compared to S is 30:60, hence P(S | D) = 60/90 = 2/3
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Unformatted text preview: (c) Let F denote fatal accident. We have 50% of (30 + 20) = 0.5 50 = 25 fatal run into train accidents, and 80% of (60 + 20) = 0.8 80 = 64 fatal struck by train accidents, hence the total is P(F) = (25 + 64) / 130 0.685 (d) (i) D and R are not mutually exclusive; they can occur together (there were 30 run-into-train accidents happened in daytime); (ii) If D and R are, we must have P(R | D) = P(R), but here P(R | D) = 30 / 90 = 1/3, while P(R) = (30 + 20) / 130 = 5/13, so D and R are not statistically independent....
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