Prob_hw2_sol - Probability and Stochastic Processes...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Probability and Stochastic Processes Homework 2 Solutions 5th Jan,Spring’2010 Problem Solutions : Yates and Goodman, 1.7.3 1.7.5 1.7.7 1.8.1 1.8.2 1.8.6 1.8.7 1.9.1 1.9.3 1.9.4 1.10.1 1.10.2 1.10.3 and 1.11.3 Problem 1.7.3 Solution Let G i and B i denote events indicating whether free throw i was good ( G i ) or bad ( B i ). The tree for the free throw experiment is a8 a8 a8 a8 a8 a8 G 1 1 / 2 a72 a72 a72 a72 a72 a72 B 1 1 / 2 a24 a24 a24 a24 a24 a24 G 2 3 / 4 a88 a88 a88 a88 a88 a88 B 2 1 / 4 a24 a24 a24 a24 a24 a24 G 2 1 / 4 a88 a88 a88 a88 a88 a88 B 2 3 / 4 G 1 G 2 3 / 8 G 1 B 2 1 / 8 B 1 G 2 1 / 8 B 1 B 2 3 / 8 The game goes into overtime if exactly one free throw is made. This event has probability P [ O ] = P [ G 1 B 2 ] + P [ B 1 G 2 ] = 1 / 8 + 1 / 8 = 1 / 4 . (1) Problem 1.7.5 Solution The P [ − | H ] is the probability that a person who has HIV tests negative for the disease. This is referred to as a false-negative result. The case where a person who does not have HIV but tests positive for the disease, is called a false-positive result and has probability P [+ | H c ]. Since the test is correct 99% of the time, P [ −| H ] = P [+ | H c ] = 0 . 01 . (1) Now the probability that a person who has tested positive for HIV actually has the disease is P [ H | +] = P [+ , H ] P [+] = P [+ , H ] P [+ , H ] + P [+ , H c ] . (2) We can use Bayes’ formula to evaluate these joint probabilities. P [ H | +] = P [+ | H ] P [ H ] P [+ | H ] P [ H ] + P [+ | H c ] P [ H c ] (3) = (0 . 99)(0 . 0002) (0 . 99)(0 . 0002) + (0 . 01)(0 . 9998) (4) = 0 . 0194 . (5) Thus, even though the test is correct 99% of the time, the probability that a random person who tests positive actually has HIV is less than 0.02. The reason this probability is so low is that the a priori probability that a person has HIV is very small. 1
Image of page 1

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

View Full Document Right Arrow Icon
Problem 1.7.7 Solution The tree for this experiment is a8 a8 a8 a8 a8 a8 A 1 1 / 2 a72 a72 a72 a72 a72 a72 B 1 1 / 2 a24 a24 a24 a24 a24 a24 H 1 1 / 4 T 1 3 / 4 H 1 3 / 4 a88 a88 a88 a88 a88 a88 T 1 1 / 4 a24 a24 a24 a24 a24 a24 H 2 3 / 4 T 2 1 / 4 H 2 3 / 4 a88 a88 a88 a88 a88 a88 T 2 1 / 4 a24 a24 a24 a24 a24 a24 H 2 1 / 4 T 2 3 / 4 H 2 1 / 4 a88 a88 a88 a88 a88 a88 T 2 3 / 4 A 1 H 1 H 2 3 / 32 A 1 H 1 T 2 1 / 32 A 1 T 1 H 2 9 / 32 A 1 T 1 T 2 3 / 32 B 1 H 1 H 2 3 / 32 B 1 H 1 T 2 9 / 32 B 1 T 1 H 2 1 / 32 B 1 T 1 T 2 3 / 32 The event H 1 H 2 that heads occurs on both flips has probability P [ H 1 H 2 ] = P [ A 1 H 1 H 2 ] + P [ B 1 H 1 H 2 ] = 6 / 32 . (1) The probability of H 1 is P [ H 1 ] = P [ A 1 H 1 H 2 ] + P [ A 1 H 1 T 2 ] + P [ B 1 H 1 H 2 ] + P [ B 1 H 1 T 2 ] = 1 / 2 . (2) Similarly, P [ H 2 ] = P [ A 1 H 1 H 2 ] + P [ A 1 T 1 H 2 ] + P [ B 1 H 1 H 2 ] + P [ B 1 T 1 H 2 ] = 1 / 2 . (3) Thus P [ H 1 H 2 ] negationslash = P [ H 1 ] P [ H 2 ], implying H 1 and H 2 are not independent. This result should not be surprising since if the first flip is heads, it is likely that coin B was picked first. In this case, the second flip is less likely to be heads since it becomes more likely that the second coin flipped was coin A .
Image of page 2
Image of page 3
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