Prob.part39.81_82 - 2.2 CONTINUOUS DENSITY For examples such as those in Exercises 9 and 10 it might seem that at least you should not have to wait

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: 2.2. CONTINUOUS DENSITY FUNCTIONS 73 11 For examples such as those in Exercises 9 and 10, it might seem that at least you should not have to wait on average more than 10 minutes if the average time between occurrences is 10 minutes. Alas, even this is not true. To see why, consider the following assumption about the times between occurrences. Assume that the time between occurrences is 3 minutes with probability .9 and 73 minutes with probability .1. Show by simulation that the average time between occurrences is 10 minutes, but that if you come upon this system at time 100, your average waiting time is more than 10 minutes. 12 Take a stick of unit length and break it into three pieces, choosing the break points at random. (The break points are assumed to be chosen simultane- ously.) What is the probability that the three pieces can be used to form a triangle? Hint : The sum of the lengths of any two pieces must exceed the length of the third, so each piece must have length < 1 / 2. Now use Exer- cise 8(g).cise 8(g)....
View Full Document

This note was uploaded on 09/15/2009 for the course SCF scf taught by Professor Scf during the Spring '09 term at Indian Institute Of Management, Ahmedabad.

Page1 / 2

Prob.part39.81_82 - 2.2 CONTINUOUS DENSITY For examples such as those in Exercises 9 and 10 it might seem that at least you should not have to wait

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