ISYE2027Hw1sol - to the 6 male positions, and similarly 6!...

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ISyE 2027B Probability with Applications Fall 2010 Homework 1 Solutions 1. (2.3) By Venn’s Diagram, P ( C c D ) = P ( D ) - P ( C D ) = 0 . 4 - 0 . 2 = 0 . 2. 2. (2.6) (a) P ( A B ) = P ( A )+ P ( B ) - P ( A B ) = 1 / 3+1 / 2 - 3 / 4 = 1 / 12. (b) By DeMorgan’s Rule, P ( A c B c ) = P (( A B ) c ) = 1 - P ( A B ) = 1 - 1 / 12 = 11 / 12. 3. (2.18) Suppose p is the probability that AT LEAST ONE lab succeeds. So p = p 2 + (1 - p ) p + p (1 - p ) = p (2 - p ). And the new failure rate is 1 - p = (1 - p ) 2 . Now the the probability to get the first success on n th day = (1 - p ) n 1 p = (1 - p ) 2( n 1) (2 - p ) p . 4. (a) Ω = { Essential Eats, Asian, Mexican, Salad Bar } . (b) Ω = { Essential Eats (vegetarian), Essential Eats (non-vegetarian), Asian(vegetarian), Asian(non-vegetarian), Mexican(vegetarian), Mexican(non- vegetarian), Salad Bar(vegetarian), Salad Bar(non-vegetarian) } . It’s OK with or without Salad Bar(non-vegetarian). 5. Problem 5. 6! × 6! = 518400. There are 6! ways of assigning the six males
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Unformatted text preview: to the 6 male positions, and similarly 6! ways of positioning the 6 females. The total number of arrangements is thus 6! 6!. The basic counting principle applies because for each arrangement of the 6 males, there are 6! ways to arrange the 6 females. 6. Problem 6. This is similar to the birthday problem. There are 15 14 13 ... 6 = 15! 5! ways of all 10 passengers getting o the bus at 10 dierent bus stops. The total number of possibilities that 10 passengers get o the bus at any of the 15 stops is 15 10 . Then the probability that no two people get o at the same bus stop equals 15 14 13 ... 6 15 10 1...
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This note was uploaded on 02/21/2011 for the course ISYE 2027 taught by Professor Zahrn during the Fall '08 term at Georgia Institute of Technology.

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