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K=29 M=9 N=1 27. There are 5 balls, numbered from 1 to 5, in a box. We draw a ball k times, with replacement. Using

K=29 M=9 N=1
27. There are 5 balls, numbered from 1 to 5, in a box. We draw a ball k times, with replacement. Using the Chebyshev-Bienaym inequality, assess an upper bound to the probability of event A - the sum of all numbers drawn will not fall into the interval (3k 3m, 3k + 3m). ANSWER: P(A) =

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K=29 M=9 N=1 27. There are 5 balls, numbered from 1 to 5, in a box. We draw a ball k times, with replacement. Using

K=29 M=9 N=1 27. There are 5 balls, numbered from 1 to 5, in a box. We draw a ball k times, with replacement. Using the Chebyshev-Bienaym inequality, assess an upper bound to the probability of event A - the sum of all numbers drawn will not fall into the interval (3k 3m, 3k + 3m). ANSWER: P(A) =

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K=29 M=9 N=1
27. There are 5 balls, numbered from 1 to 5, in a box. We draw a ball k times, with replacement. Using the Chebyshev-Bienaym inequality, assess an upper bound to the probability of event A - the sum of all numbers drawn will not fall into the interval (3k 3m, 3k + 3m). ANSWER: P(A) =