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Unformatted text preview: 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 falsenegative result. The case where a person who does not have HIV but tests positive for the disease, is called a falsepositive 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 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...
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This note was uploaded on 02/13/2011 for the course 332 226 taught by Professor Staff during the Spring '08 term at Rutgers.
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