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krzysioP - Department of Mathematicsmath.illinoisstate.edu/krzysio/2-20-10-KO-Exercise.pdfP Sample Exam Questions, Problem No. 146, also Dr. Ostaszewski’s online exercise posted

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Page 1: krzysioP - Department of Mathematicsmath.illinoisstate.edu/krzysio/2-20-10-KO-Exercise.pdfP Sample Exam Questions, Problem No. 146, also Dr. Ostaszewski’s online exercise posted

Krzys’ Ostaszewski: http://www.krzysio.net Author of the “Been There Done That!” manual for Course P/1 http://smartURL.it/krzysioP (paper) or http://smartURL.it/krzysioPe (electronic) Instructor of online P/1 seminar: http://smartURL.it/onlineactuaryIf you find these exercises valuable, please consider buying the manual or attending the seminar, and if you can’t, please consider making a donation to the Actuarial Program at Illinois State University: https://www.math.ilstu.edu/actuary/giving/ Donations will be used for scholarships for actuarial students. Donations are tax-deductible to the extent allowed by law.

If you have questions about these exercises, please send them by e-mail to: [email protected]

P Sample Exam Questions, Problem No. 146, also Dr. Ostaszewski’s online exercise posted February 20, 2010 A survey of 100 TV watchers revealed that over the last year: i) 34 watched CBS.ii) 15 watched NBC.iii) 10 watched ABC.iv) 7 watched CBS and NBC.v) 6 watched CBS and ABC.vi) 5 watched NBC and ABC.vii) 4 watched CBS, NBC, and ABC.viii) 18 watched HGTV and of these, none watched CBS, NBC, or ABC.Calculate how many of the 100 TV watchers did not watch any of the four channels(CBS, NBC, ABC or HGTV).

A. 1 B. 37 C. 45 D. 55 E. 82

Solution.We start by labeling the events. Let C be the event of a randomly chosen person watching CBS over the last year, let N be the event of watching NBC over the last year, let A be theevent of watching ABC over the last year, and finally let H the event of watching HGTVover the last year. We are given that Pr C( ) = 0.34, Pr N( ) = 0.15, Pr A( ) = 0.10,Pr C∩ N( ) = 0.07, Pr C∩ A( ) = 0.06, Pr N ∩ A( ) = 0.05, and Pr C∩ N ∩ A( ) = 0.04, aswell as Pr H( ) = 0.18, with H ∩C = H ∩ N = H ∩ A =∅. Note that

Pr C∪ N ∪ A( ) = Pr C( ) + Pr N( ) + Pr A( )− Pr C∩ N( )− Pr C∩ A( )− Pr N ∩ A( ) + Pr C∩ N ∩ A( ) == 0.34 + 0.15 + 0.10 − 0.07 − 0.06 − 0.05 + 0.04 = 0.45.

The probability we are looking for is

Page 2: krzysioP - Department of Mathematicsmath.illinoisstate.edu/krzysio/2-20-10-KO-Exercise.pdfP Sample Exam Questions, Problem No. 146, also Dr. Ostaszewski’s online exercise posted

Pr C∪ N ∪ A∪H( )C( ) = 1− Pr C∪ N ∪ A∪H( ) = 1− Pr C∪ N ∪ A( )∪HNote that C∪N∪A and Hare mutually exclusive

⎜⎜⎜

⎟⎟⎟=

= 1− Pr C∪ N ∪ A( ) + Pr H( )( ) = 1− 0.45 + 0.18( ) = 0.37.

Since there is a total of 100 persons watching TV, this means 37 people. Answer B. © Copyright 2010 by Krzysztof Ostaszewski. All rights reserved. Reproduction in whole or in part without express written permission from the author is strictly prohibited. Exercises from the past actuarial examinations are copyrighted by the Society of Actuaries and/or Casualty Actuarial Society and are used here with permission.