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Test 1, ENCI303, 23 March 2010 Formula Sheet P (A) = P (A | B1 )P (B1 ) + · · · + P (A | Bn )P (Bn ) P (A | Bi )P (Bi ) P (Bi | A) = P i P (A | Bi )P (Bi ) P (A ∪ B) = P (A) + P (B) − P (AB) V ar(X) = E(X 2 ) − (E(X))2 Binomial distribution: P (X = x) = n x px (1 − p)n−x , n x = n! x!(n−x)! Poisson distribution with rate λ: P (X = x) = e−λ λx /x! INSTRUCTIONS: 1. For two events A and B, if p = P (A | B), which of the following is always equal to 1 − p: (a) P (Ac |B c ) (b) P (Ac |B) (c) P (A|B c ) (d) P (B|A) Solution. The answer is (b), because given that B occurs, either A occurs or A doesn’t occur, and these are all the possibilities, so P (A|B)+P (Ac |B) = 1. 2. Suppose two events, A and B, are dependent, with 0 < P (A) < 1, 0 < P (B) < 1. Which of the following can be said about A and B c ? (a) A and B c are independent. (b) A and B c are mutually exclusive. 1 (c) A and B c are dependent. (d) none of the above. Solution (updated). Because A and B are dependent, P (B|A) 6= P (B). Thus, P (B c |A) = 1 − P (B|A) 6= 1 − P (B) = P (B c ) Because P (B c |A) 6= P (B c ), A and B c are not independent. 3. Suppose I flip a fair coin. If I get Heads, then I roll a 6-sided die. If I get Tails, then I roll two 6-sided dice and count the total. (a) Let X be the value of the die or dice that I get. Find P (X = 6). Solution. Use the total probability formula: P (X = 6) = P (X = 6|H)P (H) + P (X = 6|T )P (T ) = 5 1 11 1 1 · + · = 6 2 36 2 72 (b) Suppose that I get X = 6. Given that I got X = 6, what is the probability that the coin landed Heads? Solution. Use Bayes Formula. P (H|X = 6) = (1/6)(1/2) 72 6 P (X = 6|H)P (H) = = = P (X = 6) 11/72 12 · 11 11 4. Suppose the travel time between two major cities A and B by air is 6 or 7 hr if the flight is nonstop; however, if there is one stop, the travel time would be 9, 10, or 11 hr. A nonstop flight between A and B would cost $1200, whereas with one stop the cost is only $550. Then, between cities B and C, all flights are nonstop requiring 2 or 3 hours at a cost of $300. For a passenger wishing to travel from city A to city C, (a) What is the possibility space or sample space of his travel times from A to B? From A to C? Solution: S = {(6, $1200), (7, $1200), (9, $550), (10, $550), (11, $550)} (b) What is the sample space of his travel cost from A to C? 2 Solution. You can write this in a couple ways, depending on whether you keep track of just the total hours and total cost, or keep track of the outcomes for the separate portions of the flights. S = {(6, $1200, 2, $300), (7, $1200, 2, $300), (9, $550, 2, $300), (10, $550, 2, $300), (11, $550, 2, $300), (6, $1200, 3, $300), (7, $1200, 3, $300), (9, $550, 3, $300), (10, $550, 3, $300), (11, $550, 3, $300)} Or S = {(8, $1500), (9, $1500), (11, $850), (12, $850), (13, $850), (9, $1500), (10, $1500), (14, $850)} Note that some combinations can occur in more than one way. (c) If T = travel time from city A to city C, and S = cost of travel from A to C, what is the sample space of T and S? Solution. For T , the travel time has sample space S = {8, 9, 11, 12, 13, 14} and the cost has sample space S = {$850, $1500}. 5. The maximum intensity of the next earthquake in a city may be classified (for simplicity) as low (L), medium (M ), or high (H) with relative likelihoods of 15:4:1. Suppose also that buildings may be divided into two types; poorly constructed (P ) and well constructed (W ). About 20% of all the buildings in the city are known to be poorly constructed for earthquake resistance. It is estimated that a poorly constructed building will be damaged with a probability of 0.10, 0.50, or 0.90 when subjected to a low-, medium-, or highintensity earthquake, respectively. However, a well-constructed building will survive a low-intensity earthquake, although it may be damaged when subjected to a medium- or high-intensity earthquake with probability of 0.05 or 0.20, respectively. (a) What is the probability that a well-constructed building will be damaged during the next earthquake? Solution. If the relative probabilities for intensities are 15:4:1, then we have P (L) = 0.75, P (M ) = 0.2 and P (H) = 0.05. The probability of damage for a well constructed building can be found using total probability, and we assume that the intensity of the earthquake is independent of the quality of the building being considered. 3 P (D|W ) = P (D|W L)P (L|W ) + P (D|W M )P (M |W ) + P (D|W H)P (H|W ) = P (D|W L)P (L) + P (D|W M )P (M ) + P (D|W H)P (H) = 0 + (0.05)(0.2) + (0.2)(0.05) = 0.02 (b) What proportion of the buildings in this city will be damaged during the next earthquake? Solution. A similar calculation as in (a) for poorly constructed buildings gives P (D|P ) = P (D|P L)P (L) + P (D|P M )P (M ) + P (D|P H)P (H) = (0.1)(0.75) + (0.5)(0.2) + (0.9)(0.05) = 0.22 Therefore, P (D) = P (D|P )P (P ) + P (D|W )P (W ) = (0.22)(0.2) + (0.02)(0.8) = 0.06 (c) If a building in the city is damaged after an earthquake, what is the probability that the building was poorly constructed? P (P |D) = P (D|P )P (P ) (0.22)(0.2) = = 0.73 P (D) 0.06 Note that 0.73 > 0.2, the fraction of buildings that are poorly constructed, which makes sense. Knowing that a building was damaged in an earthquake makes it more likely (knowing nothing else about the building) that the building was poorly constructed. 6. The number of traffic accidents at a particular intersection has a Poisson with a rate of 1 in every 3 yr. (a) What is the probability that there will not be any accidents in the next five years? Solution. The rate over a five year interval is (1/3)(5) = 5/3, so the probability of no accidents in a five year interval is e−5/3 ≈ 0.19. 4 (b) Suppose that every accident at this intersection has a 5% chance of fatality. What is the probability that there is a fatal accident within the next 3 years? Solution. The rate of accidents in a three year period is 1. The rate of fatal accidents is (1)(0.05) = 0.05. The probability of at least one fatal accident in the next 3 years is P (X ≥ 1) = 1 − P (X = 0) = 1 − e−0.05 ≈ 0.049. 0.0 0.2 0.4 F(x) 0.6 0.8 1.0 7. Consider the cdf on the following page, which is only plotted from x = −2 to x = 7: −2 0 2 4 x Answer the following questions: (a) What is P (X = −1)? P (X = −1) = 0.2 (b) What is P (X ≤ 1)? 5 6 P (X ≤ 1) = 0.3 (c) What is P (X < 1) P (X < 1) = P (X ≤ 0) = 0.3 (d) What is P (0 < X < 2)? P (0 < X < 2) = 0 (e) Write the probability mass function P (X = i). Be sure to specify the mass function for all real values i. 0.2 0.1 0.3 P (X = i) = 0.1 0.3 0 6 i = −1 i=0 i=2 i=3 i=5 otherwise (1)