exam1sol - Rensselaer Polytechnic Institute Department of...

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Rensselaer Polytechnic Institute Department of Electrical, Computer, and Systems Engineering ECSE 2500: Engineering Probability, Fall 2011 Exam 1 Solutions The boxed numbers beside each problem give the mean (left) and median (right) score for each problem. If your score on a problem is lower than the median, you need to learn the concept better. The overall mean was 78.1 and the median was 83.5. There were 18 scores in the 90’s. 12.8 13.5 1. (15 points.) In this problem we consider two fair 4-sided dice. Die A is labeled with the numbers { 1 , 1 , 2 , 3 } and Die B is labeled with the numbers { 0 , 1 , 1 , 2 } . The two dice are rolled and we record the two results. 4.4 5 (a) (5 points.) What is the simplest and smallest sample space S we can use for this problem? How many possible outcomes are there? Since there are three outcomes for each die, the sample space S for the two results (die A, die B) has nine total outcomes: S = ±² 1 0 ³ , ² 1 1 ³ , ² 1 2 ³ , ² 2 0 ³ , ² 2 1 ³ , ² 2 2 ³ , ² 3 0 ³ , ² 3 1 ³ , ² 3 2 ³´ Since we cannot distinguish the 1’s on Die A and B, it is not correct to say we have 16 outcomes, or that we have some sort of expanded sample space with repetitions. Also, since we record the two results separately, the sample space is not the sum of the outcomes. It is critical to see that these outcomes are not all equally probable! For example, ±² 1 1 ³´ is four times as likely as ±² 3 2 ³´ . 4.4 5 (b) (5 points.) What is the event A corresponding to “The two results sum to 3”? Three of the above outcomes are in the event A : A = ±² 1 2 ³ , ² 2 1 ³ , ² 3 0 ³´ The event must be an explicit subset (a collection of outcomes) from the sample space S in part (a). While it wasn’t part of the question, P ( A ) = 5 16 . 1
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3.9 5 (c) (5 points.) Are the events B and C corresponding to “The two results sum to 2” and “Die A comes up 1” independent? Explain your reasoning. We have B = ±² 1 1 ³ , ² 2 0 ³´ P ( B ) = 5 16 C = ±² 1 0 ³ , ² 1 1 ³ , ² 1 2 ³´ P ( C ) = 1 2 B C = ±² 1 1 ³´ P ( B C ) = 1 4 Since P ( B C ) 6 = P ( B ) · P ( C ) , the two events are not independent. Another way to see this is to show that P ( B | C ) = P ( B ) or that P ( C | B ) = P ( C ) , i.e., that knowing that one event occurred doesn’t change the probability that the other event occurred. You can’t answer the question correctly without explicitly computing probabilities. Again, several people assumed that each outcome was equally likely (e.g., P ( B ) = 2 9 since it contains 2 of the 9 outcomes). This is a major misunderstanding! See also Problem 5. 2
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16.8 19 2. (20 points.) A device uses a 7-segment LED display to show a number between 0 and 9 to the user. Each of the 7 segments can be turned on or off independently, and we call a given on/off setting of all 7 segments a pattern. The ten patterns that represent valid digits are shown below, but many other patterns are possible. 4.3
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This note was uploaded on 11/29/2011 for the course ECSE 2500 taught by Professor Radke during the Spring '11 term at Rensselaer Polytechnic Institute.

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exam1sol - Rensselaer Polytechnic Institute Department of...

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