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# Hw03ans - STAT 408 Spring 2009 Homework#3(due Friday...

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STAT 408 Spring 2009 Homework #3 (due Friday, February 13, by 3:00 p.m.) Be sure to show all your work; your partial credit might depend on it. No credit will be given without supporting work. 1. A bank classifies borrowers as "high risk" or "low risk," and 16% of its loans are made to those in the "high risk" category. Of all the bank's loans, 5% are in default. It is also known that 40% of the loans in default are to high-risk borrowers. P(High risk) = 0.16, P(Default) = 0.05, P(High risk | Default) = 0.40. a) What is the probability that a randomly selected loan is in default and issued to a high-risk borrower? P(Default High risk) = P(Default) P(High risk | Default) = 0.05 0.40 = 0.02 . High Risk Low Risk Default 0.02 0.03 0.05 Default ' 0.14 0.81 0.95 0.16 0.84 1.00 b) What is the probability that a loan will default, given that it is issued to a high-risk borrower ? P(Default | High risk) = 16 . 0 02 . 0 risk) P(High risk) High P(Default = = 0.125 . c) What is the probability that a randomly selected loan is either in default or issued to a high-risk borrower, or both? P(Default High risk) = P(Default) + P(High risk) P(Default High risk) = 0.05 + 0.16 0.02 = 0.19 .

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d) A loan is being issued to a borrower who is not high-risk . What is the probability that this loan will default? 84 . 0 03 . 0 ) High risk P( ) High risk Default P( ) High risk Default P( ' ' ' | = = 0.0357 . e) Are events {a randomly selected loan is in default} and {a randomly selected loan is issued to a high-risk borrower} independent? Justify your answer . P(High risk | Default) = 0.40. P(High risk) = 0.16. Since P(High risk | Default) P(High risk), {Default} and {High risk} are NOT independent . OR P(Default | High risk) = 0.125. P(Default) = 0.05. Since P(Default | High risk) P(Default), {Default} and {High risk} are NOT independent . OR P(Default High risk) = 0.02. P(Default) × P(High risk) = 0.05 × 0.16 = 0.008. Since P(Default High risk) P(Default) × P(High risk), {Default} and {High risk} are NOT independent . 2. At Initech , 50% of all employees surf the Internet during work hours. 20% of the employees surf the Internet and play Solitaire during work hours. It is also known that 60% of the employees either surf the Internet or play Solitaire (or both) during work hours. P( Internet ) = 0.50, P( Internet Solitaire ) = 0.20, P( Internet Solitaire ) = 0.60. a) What proportion of the employees play Solitaire during work hours? P( Internet Solitaire ) = P( Internet ) + P( Solitaire ) – P( Internet Solitaire ) 0.60 = 0.50 + P( Solitaire ) – 0.20 P( Solitaire ) = 0.30 .
Solitaire Solitaire ' Internet 0.20 0.30 0.50 Internet ' 0.10 0.40 0.50 0.30 0.70 1.00 b) If it is known that an employee surfs the Internet during work hours , what is the probability that he/she also plays Solitaire ? P( Solitaire | Internet ) = 50 . 0 20 . 0 ) Internet ( P ) Internet ( P = Solitaire = 0.40 .

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Hw03ans - STAT 408 Spring 2009 Homework#3(due Friday...

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