rec02-1 - C . Note that this is a conditional version of...

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Massachusetts Institute of Technology 6.041/6.431: Probabilistic Systems Analysis (Fall 2008) Recitation 2 September 9, 2008 1. (Problem 1.14, page 56 of the text.) We roll two fair 6-sided dice. Each one of the 36 possible outcomes is assumed to be equally likely. (a) Find the probability that doubles are rolled. (b) Given that the roll results in a sum of 4 or less, find the conditional probability that doubles are rolled. (c) Find the probability that at least one die roll is a 6. (d) Given that the two dice land on different numbers, find the conditional probability that at least one die roll is a 6. 2. (Problem 1.18, page 57 of the text.) Let A Ω and B Ω be events in some sample space Ω. Show that P ( A B | B ) = P ( A | B ), assuming that P ( B ) > 0. Also show that P ( A | B ) = P ( A | B C ) P ( C | B ) + P ( A | B C c ) P ( C c | B , for any event
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Unformatted text preview: C . Note that this is a conditional version of the total probability theorem (conditioned on the event B in this case). Breaking event B up like this into the disjoint partition B C and B C c can sometimes be useful in calculating P ( A | B ). 3. (Example 1.18, page 33 of the text.) A test for a certain rare disease is assumed to be correct 95% of the time: if a person has the disease, the test results are positive with probability 0.95, and if the person does not have the disease, the test results are negative with probability 0.95. A random person drawn from a certain population has probability 0.001 of having the disease. Given that the person just tested positive, what is the probability of having the disease? Explain this result intuitively. Page 1 of 1...
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This note was uploaded on 12/02/2011 for the course ENGINEERIN EE302 taught by Professor Proasin during the Spring '11 term at South Carolina.

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