oldmidterm3sol

oldmidterm3sol - UCSD ECE153 Handout #24 Prof. Young-Han...

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: UCSD ECE153 Handout #24 Prof. Young-Han Kim Tuesday, May 3, 2011 Solutions to Midterm Examination (Spring 2010) (Prepared by TA Lele Wang) 1. Let X i denote the color of the i-th ball. (a) By symmetry, P { X 1 = R } = 1 / 2. (b) Again by symmetry, P { X i = R } = 1 / 2 for all i . Alternatively, by the law of total probability, we have P { X 2 = R } = P { X 1 = R } P { X 2 = R | X 1 = R } + P { X 1 = B } P { X 2 = R | X 1 = B } = 1 2 2 3 + 1 2 1 3 = 1 2 . (c) By the chain rule, we have P { X 1 = R,X 2 = R,X 3 = R } = P { X 1 = R } P { X 2 = R | X 1 = R } P { X 3 = R | X 2 = R,X 1 = R } = 1 2 2 3 3 4 = 1 4 . (d) Let N denote the number of red balls in the first three draws. From part (c), we know that P { N = 3 } = 1 / 4 = P { N = 0 } , where the latter identity follows by symmetry. Also we have P { N = 2 } = P { N = 1 } . Thus, P { N = 2 } must be 1 / 4. Alternatively, we have P { N = 2 } = P { X 1 = B,X 2 = R,X 3 = R } + P { X 1 = R,X 2 = B,X 3 = R } + P { X 1 = R,X 2 = R,X...
View Full Document

This note was uploaded on 09/16/2011 for the course ECE ECE153 taught by Professor Younghankim during the Spring '11 term at UCSD.

Page1 / 3

oldmidterm3sol - UCSD ECE153 Handout #24 Prof. Young-Han...

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