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Ch. 5 - Discrete Probability Distributions
Triola - Elementary Statistics 14th Edition
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Tutti i libri di testoTriola 14th EditionCh. 5 - Discrete Probability DistributionsProblema 5.4.43
Capitolo 5, Problema 5.4.43

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 you selected), while the remaining B objects are of the other type (such as lottery numbers you didn’t select), 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 objects of type B is
Mathematical formula for hypergeometric distribution calculating probabilities in sampling without replacement.
In New Jersey’s Pick 6 lottery game, a bettor selects six numbers from 1 to 49 (without repetition), and a winning six-number combination is later randomly selected. Find the probability of getting exactly four winning numbers with one ticket.

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Identify the parameters for the hypergeometric distribution based on the problem: - Total population size \(N = A + B = 49\) (numbers from 1 to 49), - Number of type A objects \(A = 6\) (the winning numbers), - Number of type B objects \(B = 49 - 6 = 43\) (non-winning numbers), - Sample size \(n = 6\) (numbers selected on the ticket), - Number of type A objects drawn \(x = 4\) (exactly four winning numbers).
Recall the hypergeometric probability formula: \(P(x) = \frac{\binom{A}{x} \cdot \binom{B}{n - x}}{\binom{A + B}{n}}\) where \(\binom{m}{k} = \frac{m!}{k!(m-k)!}\) is the binomial coefficient representing combinations.
Calculate the number of ways to choose exactly \(x=4\) winning numbers from the \(A=6\) winning numbers: \(\binom{6}{4}\).
Calculate the number of ways to choose the remaining \(n - x = 2\) numbers from the \(B=43\) non-winning numbers: \(\binom{43}{2}\).
Calculate the total number of ways to choose any \(n=6\) numbers from the \(N=49\) numbers: \(\binom{49}{6}\). Finally, plug these values into the formula to find the probability of exactly four winning numbers.

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Hypergeometric Distribution

The hypergeometric distribution models the probability of drawing a specific number of successes from a finite population without replacement. It applies when sampling is done without replacement, making events dependent. The formula calculates the probability of obtaining exactly x successes in n draws from a population with A successes and B failures.
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Introduction to the Hypergeometric Distribution

Sampling Without Replacement

Sampling without replacement means each selected item is not returned to the population before the next draw. This causes the probabilities to change after each draw, making events dependent. This contrasts with sampling with replacement, where each draw is independent and the binomial distribution applies.
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Sampling Distribution of Sample Proportion

Lottery Probability Application

In lottery problems like New Jersey’s Pick 6, the hypergeometric distribution helps calculate the chance of matching a certain number of winning numbers. Here, the population is the set of all numbers, successes are the winning numbers, and the sample is the player's chosen numbers. This real-world example illustrates the use of hypergeometric probabilities.
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Introduction to Probability
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