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# midsoln1 - 332:321 Probability and Stochastic Processes...

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332:321 Probability and Stochastic Processes Examination 1 October 14, 2004 SOLUTION (version 1) This exam was given in two versions. The questions were mostly the same but with different numeric values. Solutions are included for both versions. You have 80 minutes to answer the following four questions in the notebooks provided. You are permit- ted 1 double-sided sheet of notes. Make sure that you have included your name, Rutgers netid and signature in each book used. (5 points) Read each question carefully. All statements must be justified. Computations should be simplified as much as possible. Make sure to read both sides of this exam paper. 1. 30 points SHORT ANSWER You must explain your answer. (a) Events A and B are independent, P [ A ] = P [ B ] and P [ A B ] = P [ A ] + P [ B ]. What is P [ A ]? For any events A and B, P [ A B ] = P [ A ] + P [ B ] P [ AB ] . It follows that P [ ] = 0 . Since A and B are independent and P [ A ] = P [ B ] , P [ A ] 2 = P [ A ] P [ B ] = P [ ] = 0 . Thus P [ A ] = 0 . (b) K is a Poisson = 2 ) random variable. What is P [ K < E [ K ]]? Since K is a Poisson = 2 ) random variable, K as PMF P K ( k ) = ( 2 k e 2 / k ! k = 0 , 1 , 2 ,... 0 otherwise Since E [ K ] = 2 , P [ K < E [ K ]] = P [ K < 2] = P K ( 0 ) + P K ( 1 ) = 2 0 e 2 / 0! + 2 1 e 2 / 1! = 3 e 2 (c) On each turn of the knob, a gumball machine is equally likely to dispense a red, yellow, green or blue gumball, independent from turn to turn. After eight turns, what is the probability P [ R 2 Y 2 G 2 B 2 ] of receiving 2 red, 2 yellow, 2 green and 2 blue gumballs? In this problem, we are drawing gumball samples from the machine without replacement. The requested probability is given by the multinomial probability P [ R 2 Y 2 G 2 B 2 ] = 8! 2! 2! 2! 2! ± 1 4 ² 2 ± 1 4 ² 2 ± 1 4 ² 2 ± 1 4 ² 2 = 8! 4 10 0 . 0385 . 2. 50 points A particular birth defect of the heart is very rare; a random newborn infant will have the defect with probability P [ D ] = 10 4 . In the regular exam of a newborn, a particular heart arrhythmia A occurs with probability 0 . 99 in infants with the defect D . However, the arrhythmia A also appears with probability 0 . 1 in infants without the defect D . When the arrhythmia is present, a lab test is performed. The result of the lab test is either positive (event T + ) or negative (event T ). The lab does the test correctly with probability p = 0 . 999, independent from test to test. When the test is done correctly, the test result correctly indicates whether the infant has the defect. When the lab does the test incorrectly, the test result is always wrong. If the arrhythmia is present and the test is positive, then the heart surgery (event H ) is performed to repair the defect. 1

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(a) Without doing any calculations, find P [ A | D ], P ± T + | D ² , P ± T + | D c ² , P ± T | D ² and P ± T | D c ² Directly from the problem statement, P [ A | D ] = 0 . 99 . Unfortunately the other quantities are not so simple. The intent of problem was to use the problem statement that the test gives the correct result with probability 0 . 999 to draw the following conclusions that P ± T + | D ² = 0 . 999 P ± T | D ² = 0 . 001 P ± T + | D c ² = 0 . 001 P ± T | D c ² = 0 . 999 However, this is not precisely correct because those conditional probabilities assume that a lab test occurs. In fact, the lab test may not occur. If you draw the tree (which I unfortunately told
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## This note was uploaded on 02/13/2011 for the course 332 226 taught by Professor Staff during the Spring '08 term at Rutgers.

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midsoln1 - 332:321 Probability and Stochastic Processes...

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