Assignment 2_Solution

# Assignment 2_Solution - THE UNIVERSITY OF HONG KONG...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT 1301 PROBABILITY AND STATISTICS I Suggested Solution to Assignment 2 NB : For simplicity of notation, denote A T B = AB . 1. Solution to Question A1. (a) P ( A S B ) = P ( A ) + P ( B ) - P ( A T B ) = 0 . 55 + 0 . 65 - 0 . 3 = 0 . 9 . Similarly, P ( B S C ) = 0 . 65 + 0 . 4 - 0 . 2 = 0 . 85 and P ( A S C ) = 0 . 55 + 0 . 4 - 0 . 2 = 0 . 75. (b) P ( A S B S C ) = P (Ω) = 1; P ( A T B T C ) = P ( A S B S C ) - [ P ( A )+ P ( B )+ P ( C ) - P ( A T B ) - P ( B T C ) - P ( A T C )] = 1 - (0 . 55+0 . 65+0 . 4 - 0 . 3 - 0 . 2 - 0 . 2) = 0 . 1. (c) P ( A/B ) = P ( A T B c ) = P ( A ) - P ( AB ) = 0 . 55 - 0 . 3 = 0 . 25; P ( B/A ) = P ( B T A c ) = P ( B ) - P ( AB ) = 0 . 65 - 0 . 3 = 0 . 35. (d) P ( A | B ) = P ( A T B ) P ( B ) = 0 . 3 0 . 65 0 . 46; P ( B | A ) = P ( B T A ) P ( A ) = 0 . 3 0 . 55 = 0 . ˙ 5 ˙ 4. (e) P ( C | A/B ) = P ( C T A T B c ) P ( A/B ) = P ( C T A ) - P ( C T A T B ) P ( A/B ) = 0 . 2 - 0 . 1 0 . 25 = 0 . 4; P ( C | B/A ) = P ( C T A T B c ) P ( B/A ) = P ( C T B ) - P ( C T A T B ) P ( B/A ) = 0 . 2 - 0 . 1 0 . 35 0 . 2857; 2. Solution to Question A2. The expected overall recovery rate of drug A can be denoted as 80 · 0 . 20 + 20 · 0 . 85 100 ; 1

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Let m denote the size of the women in the treatment of drug B. The expected overall recovery rate of drug B can be denoted as (100 - m ) · 0 . 15 + m · 0 . 80 100 ; 80 · 0 . 20 + 20 · 0 . 85
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## This note was uploaded on 04/29/2010 for the course STAT 1301 taught by Professor Smslee during the Spring '08 term at HKU.

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Assignment 2_Solution - THE UNIVERSITY OF HONG KONG...

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