91Ó°ÊÓ

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 \(n-x\) objects of type \(B\) is $$P(x)=\frac{A !}{(A-x) ! x !} \cdot \frac{B !}{(B-n+x) !(n-x) !} \div \frac{(A+B) !}{(A+B-n) ! n !}$$ 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 probabilities of getting exactly two winning numbers with one ticket. (Hint: Use \(A=6, B=43, n=6\) and \(x=2 .\))

Step by step solution

01

Identify the given values

Given the problem, identify the values for the hypergeometric distribution: \(A = 6\) (winning numbers) \(B = 43\) (other numbers) \(n = 6\) (total numbers drawn) \(x = 2\) (numbers that are winning)

02

Write the hypergeometric probability formula

Here is the formula for the hypergeometric probability function: \[ P(x) = \frac{ \binom{A}{x} \cdot \binom{B}{n-x} }{ \binom{A+B}{n} } \] Where: \( \binom{a}{b} = \frac{a!}{b!(a-b)!}\)

03

Calculate the combinations

First, calculate the binomial coefficients for \(\binom{6}{2}\), \(\binom{43}{4}\), and \(\binom{49}{6}\): \[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2! \cdot 4!} = \frac{720}{2 \cdot 24} = 15 \] \[ \binom{43}{4} = \frac{43!}{4!(43-4)!} = \frac{43!}{4! \cdot 39!} = \frac{43 \cdot 42 \cdot 41 \cdot 40}{24} = 12341.5 \] \[ \binom{49}{6} = \frac{49!}{6!(49-6)!} = \frac{49!}{6! \cdot 43!} = \frac{49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44}{720} = 13983816 \]

04

Substitute the values into the hypergeometric formula

Substitute the calculated binomial coefficients into the hypergeometric probability formula: \[ P(2) = \frac{ \binom{6}{2} \cdot \binom{43}{4} }{ \binom{49}{6} } = \frac{15 \cdot 12341.5}{13983816} \]

05

Simplify the probability expression

Simplify the expression to find the probability: \[ P(2) = \frac{15 \cdot 12341.5}{13983816} = \frac{185122.5}{13983816} \approx 0.01325 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

finite population sampling

In many real-life scenarios, we deal with a finite population. This means there is a limited number of items or individuals in the group we are studying. When we take samples from this finite population, especially without replacement (not putting items back), each event can change the probability of the next event.
For example, in a lottery, once a number is drawn, it cannot be drawn again in the same draw. This makes the events dependent. Unlike the binomial distribution, which assumes independent events, the hypergeometric distribution is better suited for such situations.

probability calculation

Calculating the probability of an event is crucial in statistics. To find this probability in scenarios with dependent events (like drawing lottery numbers without replacement), we use specialized formulas.
For the hypergeometric distribution, the formula to find the probability of drawing exactly x objects of a certain type is:
\[ P(x) = \frac{ \binom{A}{x} \binom{B}{n-x} }{ \binom{A+B}{n} } \]
Here, \(\binom{a}{b}\) represents a combination, which we'll explore next. This formula helps us calculate the precise probability of specific outcomes based on our sample and population values.

combinatorial analysis

Combinatorial analysis deals with counting and arrangement. A fundamental concept here is the binomial coefficient, represented as \(\binom{a}{b}\). This coefficient counts the number of ways to choose b items from a without regard to the order. It's calculated using:
\[ \binom{a}{b} = \frac{a!}{b!(a-b)!} \]
In our example:

• Finding \( \binom{6}{2} \) gives us the number of ways to choose 2 winning numbers from 6.
• \( \binom{43}{4} \) tells us how to choose the remaining 4 numbers from the other 43.
• And \( \binom{49}{6} \) indicates the total ways to choose all 6 numbers from the 49 available.

Using these values helps us break down the problem into manageable parts, making the final probability calculation clearer.

hypergeometric probability formula

The hypergeometric probability formula is a tool for computing the likelihood of a specific number of successes in a sample drawn from a finite population. This formula is tailored for situations where events are not independent.
Given parameters from our provided exercise: \(A = 6\) (winning numbers), \(B = 43\) (other numbers), \(n = 6\) (total drawn), and \(x = 2\) (winning numbers), we substitute these into the hypergeometric formula:
\[ P(x) = \frac{ \binom{6}{2} \binom{43}{4} }{ \binom{49}{6} } \]
After computing:

• \( \binom{6}{2} = 15 \),
• \( \binom{43}{4} = 12341.5 \),
• and \( \binom{49}{6} = 13983816 \),

We find:
\[ P(2) = \frac{15 \times 12341.5}{13983816} \]
Simplifying, we get approximately 0.01325. Thus, the probability of getting exactly 2 winning numbers in the lottery with one ticket is about 1.325%.