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HW4_solutions_b - HW#4 Solutions 2011 AMS 311 Probability...

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HW#4, Solutions AMS 311 Probability Theory (Spring 2011) Part 1 (10 points): PRINT your name and StonyBrook ID ( Last, First, #ID ) on the top on your work sheet. Staple your HW if more than 1 pages. Part 2: Problems (90 points) Write down your work step by step to get full score. (1). For a given defendant, the number of votes for guilty follows a binomial distribution. So , (which means at least 9 jurors vote an innocent person “guilty”). Similarly, The event that the jury renders a wrong decision could be decomposed to disjoint events: voting a guilty person innocent, and voting an innocent person guilty. That is: , and we know P(G)=0.65. So P(E)=1-P(E)= (2). Let X a , X b , X c be the error number in this 7-page article if it is typed by Alan, Bob or Cathy, respectively. The best model is Poisson Process. That is X a ~Poisson(), =3*7=21; X b ~Poisson(), =4.2*7=29.4; X c ~Poisson(), =2.1*7=14.7. Let X be the error number in this 7-page article. Using the law of total probability, P(X=0)=P(X=0|Alan typed it)P(Alan typed it)+ P(X=0|Bob typed it)P(Bob typed it)
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