Homework 2 Solutions.pdf - ECE 3020A Homework 2 Spring 2018...

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1 ECE 3020A Homework 2, Spring 2018 Solution Problem 1. a) ( 6 2 ) + ( 7 2 ) + ( 4 2 ) = 42 possibilities. b) There are 6 × 7 choices of Impala and Malibu, 6 × 4 choices of Impala and Corvette, and 7 × 4 choices of Malibu and Corvette. That is, there are 94 possibilities. Problem 2. Let us call r for right and u for up. Then, each path is a linear arrangement of 4 r’ s and 3 u ’s. For example: r,r,u,u,r,r,u . As such, the number of possible different paths from A to B is : 7! 3!×4! = 35 . Problem 3. Let G denote the event that th e suspect “S” is guilty and C the event that he possesses the characteristic of the criminal, we have: 𝑃(𝐺|𝐶) = 𝑃(𝐺 ∩ 𝐶) 𝑃𝐶 = 𝑃(𝐶|𝐺)𝑃(𝐺) 𝑃(𝐶|𝐺)𝑃𝐺 + 𝑃(𝐶|𝐺 𝑐 )𝑃(𝐺 𝑐 ) = 1 × (0.6) 1 × (0.6) + (0.2) × .04 ≈ 0.882 Problem 4. We’ll compute each measure by its definition, using the fact that each coin flip is fair and independent: First, observe that Player 1 wins a dollar if at least 1 of the coins lands on heads.
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