# 1 - 1 A Question 1 p=0.8 prize=100 \$ B Question 2 p=0.5...

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

A Question 1 p=0.8, prize=100 \$ B Question 2 p=0.5, prize= 200 \$ E[A] = 0*0.2 + 100*(0.8*0.5) + 300*(0.8*0.5) = 0 + 40 + 120 = 160 E[B]= 0*0.5 + 200*(0.5*0.2) + 300*(0.5*0.8) = 0 + 20 + 120 =140 As we see the expected value of A is greater than B. So first question should be chosen. 2) E: event that the man discovers the matchbox in his right pocket is empty and there are k matches in the matchbox in his left pocket. If the (N+1)th choice of the matchbox in his right pocket is made at the (N+1+N-k) trial, this event will occur. E is a random variable with the negative binomial distribution p=1/2 , r= N+1, n=2N-k+1 = - ( ) - + PE 2N kN 12 2N k 1 Also if we think the opposite of this event, which we can say that being empty of the matchbox in his left pocket, the probability will be equally likely as; = - ( ) - PE 2N kN 12 2N k 3) i (number of the winnings) p (the probability that A team wins) E[X] (expected number of the games) a) i=2 There can be totally 2 games or 3 games in this case. E[X] = (2*1! )*(p

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.

## This note was uploaded on 07/30/2011 for the course END 331 taught by Professor Ilkertopcu during the Spring '11 term at Istanbul Technical University.

### Page1 / 4

1 - 1 A Question 1 p=0.8 prize=100 \$ B Question 2 p=0.5...

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

View Full Document
Ask a homework question - tutors are online