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Unformatted text preview: IE 4521 Practice Midterm Problems John Gunnar Carlsson October 24, 2009 1 Probability problems 1.1 Dice Suppose we roll a pair of dice: 1. What is the probability that the second die lands on a higher value than the first die? Solution By writing out all pairs as in slide 4 of lecture 2, we find that the second die is higher than the first die in 15 of the 36 cases, so the answer is 15/36 = 5/12. 2. What is the probability that the sum of the upturned faces is i , for i ∈ { 2, . . . , 12 } ? Solution See slide 6 of lecture 2. 3. Suppose that we know the sum of the upturned faces is either 5 or 7. What is the probability that the sum is 5? Solution Let A be the event that the sum of the faces is 5 and B the event that the sum of the faces is 7. Let C = A ∪ B be the event that the sum of the faces is either 5 or 7. Then Pr ( A ) = 4/36, Pr ( B ) = 6/36, and Pr ( C ) = Pr ( A ) + Pr ( B ) = 10/36. Then Pr ( A  C ) = Pr ( A ∩ C ) Pr ( C ) = Pr ( A ) Pr ( C ) = 4/36 10/36 = 2/5 1.2 Machine failure A system is composed of 5 components, each of which is either working or broken. Consider an experiment that consists of observing the status of each component, and let the outcome of the experiment be given by the vector ( x 1 , x 2 , x 3 , x 4 , x 5 ) , where x i = 1 if component i works and 0 if component i is broken. 1. How many outcomes are in the sample space of this experiment? Solution There are 2 5 = 32 states. 2. Suppose that the system will work if components 1 and 2 are both working, or if components 3 and 4 are both working, or if components 1, 3, and 5 are all working. Let W denote the event that the system works. Specify all the outcomes in W . Solution The event W consists of all events of the form ( 1, 1, x 3 , x 4 , x 5 ) , ( x 1 , x 2 , 1, 1, x 5 ) , and ( 1, x 2 , 1, x 4 , 1 ) . There are 15 total events which can be computed by writing all the possibilities out exhaustively. 3. Let A denote the event that components 4 and 5 are broken. How many outcomes are contained in event A ? 1 Solution Event A consists of all outcomes of the form ( x 1 , x 2 , x 3 , 0, 0 ) . Since we can freely choose x 1 , x 2 , and x 3 , there are 2 3 = 8 outcomes. 4. How many outcomes are contained in event A ∩ W ? Solution By enumerating possibilities (i.e. just writing them all out), we find that outcomes ( 1, 1, 1, 0, 0 ) and ( 1, 1, 0, 0, 0 ) are common to A and W . 2 Conditional probability problems 2.1 Witness reliability You are a member of a jury judging a hitandrun driving case. A taxi hit a pedestrian one night and fled the scene. The entire case against the taxi company rests on the evidence of one witness who saw the accident from his window some distance away. He says that he saw the pedestrian struck by a blue taxi. The lawyer for the injured pedestrian establishes the following facts: 1. There are only two taxi colors in town, blue taxis and black taxis. On the night in question, 85% of all taxis on the road were black and 15% were blue.on the road were black and 15% were blue....
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This note was uploaded on 02/19/2011 for the course IE 4521 taught by Professor Staff during the Spring '08 term at Minnesota.
 Spring '08
 Staff

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