STAT 400 hw2_ans

# STAT 400 hw2_ans - Homework#2(10 points(due Friday Feb 3 by...

This preview shows pages 1–4. Sign up to view the full content.

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework #2 (10 points) (due Friday, Feb 3, by 3:00 p.m.) No credit will be given without supporting work. 1. Bob is applying for a job with five companies. At the first company, he is in the final group of four applicants, one of which will be chosen for the position. At two of the five companies, Bob is one of ten candidates; and at the last two companies, he is in an early stage of application in a pool of 25 candidates. Assuming that all companies make their decisions independently of each other, and that Bob is as likely to be chosen as any other applicant, what is the probability of getting at least one job offer? “at least one” = “either 1st or 2nd or 3rd or 4th or 5th” = union. P(at least one) = 1 P(none). “none” = “not 1st and not 2nd and not 3rd and not 4th and not 5th”. P(A 1 A 2 A 3 A 4 A 5 ) = ) A A A A P(A 1 ' ' ' ' ' 5 4 3 2 1 since the companies are independent = ) A ( P ) A ( P ) A ( P ) A ( P ) P(A 1 ' ' ' ' ' 5 4 3 2 1 = 1 (0.75) (0.90) (0.90) (0.96) (0.96) = 1 0.56 = 0.44 . OR P(at least one job offer) = 0.25 + ( 0.75 0.10 ) + ( 0.75 0.90 0.10 ) + ( 0.75 0.90 0.90 0.04 ) + ( 0.75 0.90 0.90 0.96 0.04 ) = 0.44 . OR P(A 1 A 2 A 3 A 4 A 5 ) = P(A 1 ) + P(A 2 ) + P(A 3 ) + P(A 4 ) + P(A 5 ) – P(A 1 A 2 ) – P(A 1 A 3 ) – P(A 1 A 4 ) – P(A 1 A 5 ) – P(A 2 A 3 ) STAT 400 Spring 2012 – P(A 2 A 4 ) – P(A 2 A 5 ) – P(A 3 A 4 ) – P(A 3 A 5 ) – P(A 4 A 5 ) + P(A 1 A 2 A 3 ) + P(A 1 A 2 A 4 ) + P(A 1 A 2 A 5 ) + P(A 1 A 3 A 4 ) + P(A 1 A 3 A 5 ) + P(A 1 A 4 A 5 ) + P(A 2 A 3 A 4 ) + P(A 2 A 3 A 5 ) + P(A 2 A 4 A 5 ) + P(A 3 A 4 A 5 ) – P(A 1 A 2 A 3 A 4 ) – P(A 1 A 2 A 3 A 5 ) – P(A 1 A 2 A 4 A 5 ) – P(A 1 A 3 A 4 A 5 ) – P(A 2 A 3 A 4 A 5 + P(A 1 A 2 A 3 A 4 A 5 ) = … 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 ....
View Full Document

## This note was uploaded on 03/14/2012 for the course STAT 400 taught by Professor Kim during the Spring '08 term at University of Illinois, Urbana Champaign.

### Page1 / 12

STAT 400 hw2_ans - Homework#2(10 points(due Friday Feb 3 by...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online