HW2_solutions_b(1)

HW2_solutions_b(1) - HW#2 Solutions AMS 311 Probability...

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HW#2 Solutions AMS 311 Probability Theory (Spring 2011) Read: Ross, sections 3.1-3.4 of Ch3. You’d better go through the following examples before doing the assignments: Ch3: 2a-2g; 3a-3g; 3i-3n; 4a-4f 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) (5 points) To find P(at least one 6 | sum at least 15): {Sum at least 15}={(3,6,6); (4,5,6);(4,6,6);(5,5,5);(5,5,6);(5,6,6);(6,6,6)}, notice (4,5,6) has 3!=6 different arrangements, (3,6,6), (5,5,6) and (5,6,6) have 3 different arrangements, so #{Sum at least 15}=6+1+3+3+3+3+1=20; P(at least one 6 | sum at least 15)=P(at least one 6 and sum at least 15)/P(sum at least 15)=19/20 (2). (10 points) P(Male)=0.72, P(Female)=0.28; P(Success|Male)=0.17, P(Success|Female)=0.41 (a) To find P(Female|Success) Applying Bayes’s Theorem, P(Female|Success)= P(Female)*P(Success|Female)/[P(Female)*P(Success|Female)+P(Male)*P(Success|Male)] =0.28*0.41/(0.28*0.41+0.72*0.17)=0.4840 (b) P(attending party)=P(Success)=P(Female)*P(Success|Female)+P(Male)*P(Success|Male) =0.28*0.41+0.72*0.17=0.2372 (3). (12 points) (a) P(1 st =3 and 2 nd >4)=0.1*6/9=1/15; (b) P(both <3)=; (c) P(both>3)= Homework is due at the beginning of class on its due date. You must be in class and on time to submit your homework. No homework will be accepted via email. No late homework will be accepted. Your lowest homework grade will be dropped before computing your average.
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This note was uploaded on 05/09/2011 for the course AMS 311 taught by Professor Tucker,a during the Spring '08 term at SUNY Stony Brook.

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HW2_solutions_b(1) - HW#2 Solutions AMS 311 Probability...

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