Hw1-sol - AMS 311(Fall 2010 Joe Mitchell PROBABILITY THEORY Homework Set 1 Solution Notes(1(18 points There are 6 people at the security check at

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: AMS 311 (Fall, 2010) Joe Mitchell PROBABILITY THEORY Homework Set # 1 Solution Notes (1). (18 points) There are 6 people at the security check at JFK, each with a laptop. Unfortunately, all the laptops look identical. They put them through the machine and each person grabs a laptop at random on the other side, not noticing that they may not have grabbed their own. (a). Describe the sample space corresponding to this experiment. (Make certain to define the notation you use to specify the outcomes!) Let the people be labelled 1, 2, 3, 4, 5, 6. We consider an outcome to be a spelling using the digits 123456, where, e.g., 231465 means that laptop 1 goes to person 2, laptop 2 goes to person 3, laptop 3 goes to person 1, laptop 4 goes to person 4, laptop 5 goes to person 6, and laptop 6 goes to person 5. (The i th digit says which person gets laptop i ; alternatively, one can define the outcome by saying the i th digit says which laptop goes to person i .) Then S is the set of all 6! = 720 spellings (permutations) of the digits 123456. (b). Find the probability that each person gets her/his own laptop. If E is the event that each person gets her/his own laptop, then E = { 123456 } , consisting just of the one outcome 123456. Thus, P ( E ) = 1 / 6! = 1 / 720. (c). Find the probability that Joe (one of the 6 travelers) gets his own laptop. We can assume that Joe is person 1. If F is the event that Joe gets his own laptop, then F = { 123456 , 124356 , 136245 , 134625 , 146235 ,... } , where the set includes all 5! = 120 spellings of 123456 that begin with 1. Thus, P ( F ) = 120 / 6! = 120 / 720 = 1 / 6. (2). (20 points) Suppose that A and B are mutually exclusive events for which P ( A ) = 0 . 35 and P ( B ) = 0 . 51 . Suppose further that B and C are mutually exclusive events, and that P ( C ) = 0 . 4 ....
View Full Document

This note was uploaded on 07/15/2011 for the course AMS 311 taught by Professor Tucker,a during the Spring '08 term at SUNY Stony Brook.

Page1 / 3

Hw1-sol - AMS 311(Fall 2010 Joe Mitchell PROBABILITY THEORY Homework Set 1 Solution Notes(1(18 points There are 6 people at the security check at

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online