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oldmidterm3sol

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

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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 3 = B } = P { X 1 = B } P { X 2 = R | X 1 = B } P { X 3 = R | X 2 = R, X 1 = B } + P { X 1 = R } P { X 2 = B | X 1 = R } P { X 3 = R | X 2 = B, X 1 = R } + P { X 1 = R } P { X 2 = R | X 1 = R } P { X 3 = B | X 2 = R, X 1 = R } = 1 2 × 1 3 × 2 4 + 1 2 × 1 3 × 2 4 + 1 2 × 2 3 × 1 4 = 1 4 .

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