SolutionHW2-SP11

# SolutionHW2-SP11 - Remark independence assumption is highly...

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

STAT 427 Home Work Assignment #2 Spring 2011 Dr. Goel Solution REMARK: In part c,   1 2 3 1 2 3 1 2 3 P exactly one ( ' ') ( ' ') ( ' ' ) P A A A P A A A P A A A

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

View Full Document
E. P(M| SS)= P(M and SS)/P(SS) =(.08+.07+.12)/( .08+0.07+0.12+0.04+0.02+0.05+0.03+0.07+0.08) =0.27/0.56=0.50 F. P(SS|M) = P(SS and M)/P(M) = (.08+.07+.12)/(.08+.07+.12+0.1+0.05+0.07) = 0.27/0.49 = 0.55 Now, P(LS|M) = 1-P(SS|M )= 1-0.55 = 0.45
2.76 P(no error in all questions)=[1-P(error in one question)]^10=0.9^10=0.349 P(at least one error)=1-P(no error)=1-0.349=0.651 For general case, P(no error)=(1-p)^n and P(at least one error)=1-(1-p)^n

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: Remark: independence assumption is highly important here because it enables us to reduce the probability of intersections of events to multiplication of individual probability. 2.80 2.97 If there is impurity, then the probability that 2 out of 3 experiments are detected is: 3*0.2*0.8^2=0.384 If there is no impurity, the probability that 2 out of 3 experiments are detected is: 3*0.1^2*0.9=0.027 So the posterior probability is: 0.384*0.4/(0.384*0.4+0.027*0.6)=0.9045...
View Full Document

## This document was uploaded on 07/26/2011.

### Page1 / 4

SolutionHW2-SP11 - Remark independence assumption is highly...

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

View Full Document
Ask a homework question - tutors are online