Lecture 3 - Problem 2-105 (page 83) Illustration of...

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Problem 2-105 (page 83) Illustration of probability concepts with independent events Problem 2-108 (page 84) Illustration of probability concepts with independent events
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2-105 a) n=10 883 . 0 365 356 365 363 365 364 365 365 = 117 . 0 883 . 0 1 = b) K such that 0.5 probability that at least two have the same birthday k 23
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P( all 10 have different last 3 digits) 956 . 0 100 991 1000 999 1000 1000 = = Thus, the probability that at least two have the same last 3 digits is 1-.956=.044 The answer to part c is therefore: .117+.044-(.117)(.044)=.156
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B S 1 23 4 5
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2-108 a) 0.5 4 = 0.0625 = P(walks on 4 pitches) b) # combinations of 2 strikes on 5 pitches 10 2 5 = Total of possibilities 2 5 = 32 156 . 0 5 . 0 32 10 = (2 strikes on 5 pitches followed by a ball on the 6 th pitch)
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c) P( walk on 4) + P( walk on 5) + P( walk on 6) 4 4 0.0625 [ / 2 ]0.5 0.156 0.3435 1 ⎛⎞ =+ + = ⎜⎟ ⎝⎠ d) = P( first batter scores before first out) = 0.3435 4 0139 . 0
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Problem 2-36 (page 66) Illustrates Permutations and Combinations
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2-36 Candidate A 3 votes Candidate B 2 votes Draw the slips of paper at random. What is the probability that candidate A remains ahead of candidate B through the vote count?
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This note was uploaded on 04/08/2008 for the course ENGR 2600 taught by Professor Malmborg during the Spring '08 term at Rensselaer Polytechnic Institute.

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Lecture 3 - Problem 2-105 (page 83) Illustration of...

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