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TPR3411 Pattern Recognition Tutorial 2 (Solution) 1. The probability that Sam parks in a no-parking zone and gets police summon is 0.06. The probability that Sam cannot find a legal parking space and has to park in the no-parking zone is 0.20. On Tuesday, Sam arrives at school and has to park in a no-parking zone. Find the probability that he will get a police summon. Solution: Let N = parking in a no-parking zone S = getting a police summon Then P( S N ) 0.06 P( S | N ) 0.30 P( N ) 0.20 2. Probability that John passes a Math exam is 4/5 and that he passes a Computer programming exam is 5/6. If the probability that he passes both exams is 3/4, find the probability that he will pass at least one exam. Solution: Let M = John passes Math exam and C = John passes Computer programming exam. Then P (John passes at least one exam) = P(M U C ) = P(M) + P(C) – P(M ∩ C) = 4/5+5/6 – 3/4 = 53/60 3. For the experiment in which the number of pumps in use at a single eight-pump gas station is observed. Let A = {0, 1, 2, 3, 4, 5}, B ={4, 5, 6, 7, 8} and C = {1, 6, 7}. Then find a) A∩B b) A U B c) A U C and d) (A U C)’ Solutions: a) b) c) d) A∩B = {0, 1, 2, 3, 4, 5} ∩ {4, 5, 6, 7, 8} = {4, 5} A U B = {0, 1, 2, 3, 4, 5} U {4, 5, 6, 7, 8} = {0, 1, 2, 3, 4, 5, 6, 7, 8} = S A U C = {0, 1, 2, 3, 4, 5} U {1, 6, 7} = {0, 1, 2, 3, 4, 5, 6, 7} (A U C)’ = S - (A U C) = {0, 1, 2, 3, 4, 5, 6, 7, 8} - {0, 1, 2, 3, 4, 5, 6, 7} = {8} 4. Imagine that you went to a friend’s wedding in Australia recently. It is known that 1 in 200 people who visited Australia recently come back with swine flu. The doctor selects you at random to have a blood test for swine flu. The test is 99% accurate and the probability of a false positive is 2%. You test positive. What is the probability that you have swine flu? Solution: P( Flu ) 0.005 , P(Flu ) 0.995 P( Pos | Flu ) 0.99 , P( Neg | Flu ) 0.01 P( Pos | Flu ) 0.02 , P( Neg | Flu ) 0.98 P( Flu | Pos) P( Pos | Flu ) P( Flu ) P( Pos | Flu ) P( Flu ) P( Pos | Flu ) P(Flu ) 0.99 0.005) (0.99 0.005) (0.02 0.995) = 0.07 5. Assume that you are on a quiz show. There is a prize behind one of the two doors. The doors are coloured red and blue. A coin will be tossed to decide which door to open. You were told that there is prize behind the red door 10% of the time, and the blue door 20% of the time. If given a choice to make, which door would you open in order to win the prize? Solution: P(Red) 0.5 , P( Blue) 0.5 P(Prize | Red) 0.1 , P(Prize | Red) 0.9 P(Prize | Blue) 0.2 , P(Prize | Blue) 0.8 P(Red | Prize) P(Prize | Red) P(Red) P(Prize | Red) P(Red) P(Prize | Blue) P(Blue) 0.1 0.5 (0.1 0.5) (0.2 0.5) = 0.33 P(Blue | Prize) 1 0.33 0.67 Therefore, I would choose to open the blue door. 6. During the summer months, a retailer keeps track of the number of iPads it sells each day during a period of 90 days. The number of iPad sold per day is represented by the variable X. Compute the probability P(X) for each X. X Number of days 0 45 1 30 2 15 Total 90 Solutions: For 0 iPad: 45/90 = 0.5 For 1 iPad: 30/90 = 0.33 For 2 iPads: 15/90 = 0.17 Number of iPads sold X Probability P(X) 0 1 2 0.50 0.33 0.17 7. Find the probability of getting a 3 or 4 or 5 while throwing a die where sample spaces S={1,2,3,4,5,6,7,8,9}. Solution: Sample Spaces S={1,2,3,4,5,6,7,8,9} and event E={3,4,5}. We have, number of outcomes=3 and total number of outcomes = 9 So, P(E)=3/9=0.3333 8. The probability distribution shown in the given table represents the number holidays that an MMU student spends in the home town per 2 weeks. (That is, 18% do not go to the home town– so sad, 34% spend 1 day, 23% spends 2 days and so on). a) Find the mean b) Compute variance and standard deviation Number of days X Probability P(X) 0 1 2 3 4 0.18 0.34 0.23 0.21 0.04 Solutions: a) X P( X ) = (0)(0.18)+(1)(0.34)+(2)(0.23)+(3)(0.21)+(4)(0.04) = 1.6 b) 2 ( X ) 2 P( X ) [(0 1.6)2 0.18] [(1 1.6)2 0.34] [(2 1.6)2 0.23] [(3 1.6)2 0.21] [(4 1.6)2 0.04] = 1.23 The standard deviation is 2 1.23 1.1