practice1_sol - ORIE 3500/5500, Fall 10 Solutions for...

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: ORIE 3500/5500, Fall 10 Solutions for Practice Problems Solutions for Selected Practice Problems Problem 1 Let X A be the number of times coin A comes up heads, and let X B be the number of times coin B comes up heads. Let S A and S B denote the events that coin A or coin B is biased, respectively. We want to compute P ( S A | X A > X B ). By Bayes formula, P ( S A | X A > X B ) = P ( X A > X B | S A ) P ( S A ) P ( X A > X B | S A ) P ( S A ) + P ( X A > X B | S B ) P ( S B ) . We have P ( X A > X B | S A ) = P ( X A = 1 ,X B = 0 | S A ) + P ( X A = 2 ,X B = 0 | S A ) + P ( X A = 2 ,X B = 1 | S A ) = (2 . 7 . 3 . 5 2 ) + (0 . 7 2 . 5 2 ) + (0 . 7 2 2 . 5 2 ) = 0 . 4725 . Similarly, P ( X A > X B | S B ) = P ( X A = 1 ,X B = 0 | S B ) + P ( X A = 2 ,X B = 0 | S B ) + P ( X A = 2 ,X B = 1 | S B ) = (2 . 5 2 . 3 2 ) + (0 . 5 2 . 3 2 ) + (0 . 5 2 2 . 7 . 3) = 0 . 1725 . Also, since we have no information which coin is biased, P ( S A ) = P ( S B ) = 1 / 2 . It follows that P ( S A | X A > X B ) = (0 . 4725 . 5) (0 . 4725 . 5) + (0 . 1725 . 5) . 732 ....
View Full Document

Page1 / 5

practice1_sol - ORIE 3500/5500, Fall 10 Solutions for...

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