# Probability and Statistics for Engineers and Scientists 3e by Anthony Hayter

By Anthony Hayter

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Example text

See the previous problem. The number of ways that six people can sit around a dinner table is 5! = 120. 13 Consider 5 blocks, one block being Andrea and Scott and the other four blocks being the other four people. At the cinema these 5 blocks can be arranged in 5! ways, and then Andrea and Scott can be arranged in two different ways within their block, so that the total number of seating arrangements is 2 × 5! = 240. Similarly, the total number of seating arrangements at the dinner table is 2×4!

59. 09. 26. 323. 7817 48 CHAPTER 1. 21. 3 xi 1 2 3 4 5 6 8 9 10 pi 1 36 2 36 2 36 3 36 2 36 4 36 2 36 1 36 2 36 F (xi ) 1 36 3 36 5 36 8 36 10 36 14 36 16 36 17 36 19 36 49 50 CHAPTER 2. 000 (c) The value x = 0 is the most likely. 000 Again, x = 0 is the most likely value. 1. 6 xi -5 -4 -3 -2 -1 0 1 2 3 4 6 8 10 12 pi 1 36 1 36 2 36 2 36 3 36 3 36 2 36 5 36 1 36 4 36 3 36 3 36 3 36 3 36 F (xi ) 1 36 2 36 4 36 6 36 9 36 12 36 14 36 19 36 20 36 24 36 27 36 30 36 33 36 1 (a) xi -6 -4 -2 0 2 4 6 pi 1 8 1 8 1 8 2 8 1 8 1 8 1 8 xi -6 -4 -2 0 2 4 6 F (xi ) 1 8 2 8 3 8 5 8 6 8 7 8 1 (b) (c) The most likely value is x = 0.

27 45 (a) Let M be the event ‘male’, let E be the event ‘mechanical engineer’, and let S be the event ‘senior’. 373. 224 (a) Let T be the event that ‘the tax form is filed on time’, let S be the event that ‘the tax form is from a small business’, and let A be the event that ‘the tax form is accurate’. 13 = 24 . 022 46 CHAPTER 1. 31 The possible outcomes are (6, 5, 4, 3, 2), (6, 5, 4, 3, 1), (6, 5, 4, 2, 1), (6, 5, 3, 2, 1), (6, 4, 3, 2, 1), and (5, 4, 3, 2, 1). 32 = 1 64 = 1 65 so that the required probability is 1 1296 .