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

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern