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: Hypergeometric Distribution - If we sample from a small

Elementary Statistics | 12th Edition | ISBN: 9780321836960 | Authors: Mario F. Triola ISBN: 9780321836960 18

Solution for problem 47BB Chapter 5.3

Elementary Statistics | 12th Edition

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Elementary Statistics | 12th Edition | ISBN: 9780321836960 | Authors: Mario F. Triola

Elementary Statistics | 12th Edition

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Problem 47BB

47BB: Hypergeometric Distribution - If we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution. If a population has A objects of one type (such as lottery numbers matching the ones you selected), while the remaining ? objects are of the other type (such as lottery numbers different from the ones you selected), and if n objects are sampled without replacement (such as six drawn lottery numbers), then the probability of getting x objects of type A and n – x objects of type ? is In Pennsylvania’s Match 6 Lotto game, a bettor selects six numbers from 1 to 49 (without repetition), and a winning six-number combination is later randomly selected. Find the probabilities of the following events and express them in decimal form. a. You purchase one ticket with a six-number combination and you get exactly four winning numbers. (Hint: Use A = 6, ? = 43, n = 6, and x = 4.) b. You purchase one ticket with a six-number combination and you get all six of the winning numbers. c. You purchase one ticket with a six-number combination and you get none of the winning numbers.

Step-by-Step Solution:

Solution 47BB: The goal is to answer questions (a), (b), (c). Step 1 of 3: (a) The probability of the event “you purchase one ticket with a six-number combination and you get exactly four winning numbers” (that is, A = 6, B = 496 = 43, n = 6, x = 4) is given by: P(4) = 6! · 43! ÷ (6+43)! (64)!4! (436+4)!(64)! (6+436)!6! 6×5 43×42 6×5×4×3×2×1 = 2×1· 2×1 · 49×48×47×46×45×44 0.000969 Step 2 of 3: (b) The probability of the event “you purchase one ticket with a six-number combination and you get all six winning numbers” (that is, A = 6, B = 496 = 43, n = 6, x = 6) is given by: (6+43)! P(6) = (66)!6!·(436+6)!(66)! (6+436)!6! 6×5×4×3×2×1 = 49×48×47×46×45×44 0.0000000715

Step 3 of 3

Chapter 5.3, Problem 47BB is Solved
Textbook: Elementary Statistics
Edition: 12
Author: Mario F. Triola
ISBN: 9780321836960

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