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ISYE2027Hw1sol

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 oﬀ the bus at 10 diﬀerent bus stops. The total number of possibilities that 10 passengers get oﬀ the bus at any of the 15 stops is 15 10 . Then the probability that no two people get oﬀ at the same bus stop equals 15 × 14 × 13 × ... × 6 15 10 1...
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