problem20_36

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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• PresidentHackerCaribou10582
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20.36: For those with a knowledge of elementary probability, all of the results for this exercise are obtained from ( 29 , 2 1 )! 4 ( ! ! 4 ) 1 ( ) ( 4 - = - = - k k p p k P k n k n k where P(k) is the probability of obtaining k heads, 2 1 1 and 4 = - = = p p n for a fair coin. This is of course consistent with Fig. (20.18). a) ( 29 ( 29 16 1 4 ! 4 ! 0 ! 4 4 ! 0 ! 4 ! 4 2 1 2 1 = = for all heads or all tails.
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Unformatted text preview: b) ( 29 . 2 1 4 1 ! 3 ! 1 ! 4 4 ! 1 ! 3 ! 4 = = c) ( 29 . 2 1 8 3 4 2!2! 4! = d) . 1 2 2 8 3 4 1 16 1 = + × + × The number of heads must be one of 0, 1, 2, 3 or 4, and there must be unit probability of one and only one of these possibilities....
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• NoProfessor
• Group Theory, Probability theory, First-order logic, Elementary Probability

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