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, while the remaining B objects are of the other type, and if n objects are sampled without replacement, then the probability of getting x objects of type A and n-x objects of type B under the hypergeometric distribution is given by the following formula. In a lottery game, a bettor selects five numbers from 1 to 58 (without repetition), and a winning five-number combination is later randomly selected. Find the probabilities of getting exactly three winning numbers with one ticket. (Hint: Use A = 5, B = 53, n = 5, and x = 3.) P(x) = A! B! (A-x)!x! (B-n+x)!(n-x)! (A + B)! (A+B-n)!n! P(3)= (Round to four decimal places as needed.)
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, while the remaining B objects are of the other type, and if n objects are sampled without replacement, then the probability of getting x objects of type A and n-x objects of type B under the hypergeometric distribution is given by the following formula. In a lottery game, a bettor selects five numbers from 1 to 58 (without repetition), and a winning five-number combination is later randomly selected. Find the probabilities of getting exactly three winning numbers with one ticket. (Hint: Use A = 5, B = 53, n = 5, and x = 3.) P(x) = A! B! (A-x)!x! (B-n+x)!(n-x)! (A + B)! (A+B-n)!n! P(3)= (Round to four decimal places as needed.)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![### Understanding the Hypergeometric Distribution
When sampling from a small finite population without replacement, the binomial distribution is not appropriate because the events are not independent. Instead, the hypergeometric distribution is used when the outcomes belong to one of two types.
#### Key Concept:
If a population consists of \( A \) objects of one type and \( B \) objects of another type, and \( n \) objects are sampled without replacement, the probability of obtaining \( x \) objects of type \( A \) and \( n - x \) objects of type \( B \) is calculated using the hypergeometric distribution.
#### Hypergeometric Distribution Formula:
The formula for the hypergeometric distribution is:
\[ P(x) = \frac{A!}{(A - x)!x!} \boldsymbol{\cdot} \frac{B!}{(B - n + x)!(n - x)!} \boldsymbol{+} \frac{(A + B)!}{(A + B - n)!n!} \]
#### Application Example:
Consider a lottery scenario where a bettor selects five numbers from 1 to 58 (without repetition). A winning five-number combination is then randomly selected. We are to find the probability of getting exactly three winning numbers.
**Given Values:**
- \( A = 5 \)
- \( B = 53 \)
- \( n = 5 \)
- \( x = 3 \)
### Calculation:
#### Step-by-Step Calculation of \( P(3) \):
To calculate the probability \( P(3) \), where exactly three numbers match the winning combination, we substitute the values into the hypergeometric formula.
1. **Compute the necessary factorials:**
- \( A! \)
- \( (A - x)! \)
- \( x! \)
- \( B! \)
- \( (B - n + x)! \)
- \( (n - x)! \)
- \( (A + B)! \)
- \( (A + B - n)! \)
- \( n! \)
2. **Substitute these values into the formula** and compute the result.
Finally, the output should provide the probability, rounded to four decimal places:
\[ P(3) = \boxed{} \]
**Note:** The box indicates where the calculated probability should be filled in.
### Probabilities](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F49a5c0da-d10b-4c66-9f47-a76740d10419%2F8b180dc2-fc42-4088-b7ee-7b01a98108f3%2Fhc9o0iu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding the Hypergeometric Distribution
When sampling from a small finite population without replacement, the binomial distribution is not appropriate because the events are not independent. Instead, the hypergeometric distribution is used when the outcomes belong to one of two types.
#### Key Concept:
If a population consists of \( A \) objects of one type and \( B \) objects of another type, and \( n \) objects are sampled without replacement, the probability of obtaining \( x \) objects of type \( A \) and \( n - x \) objects of type \( B \) is calculated using the hypergeometric distribution.
#### Hypergeometric Distribution Formula:
The formula for the hypergeometric distribution is:
\[ P(x) = \frac{A!}{(A - x)!x!} \boldsymbol{\cdot} \frac{B!}{(B - n + x)!(n - x)!} \boldsymbol{+} \frac{(A + B)!}{(A + B - n)!n!} \]
#### Application Example:
Consider a lottery scenario where a bettor selects five numbers from 1 to 58 (without repetition). A winning five-number combination is then randomly selected. We are to find the probability of getting exactly three winning numbers.
**Given Values:**
- \( A = 5 \)
- \( B = 53 \)
- \( n = 5 \)
- \( x = 3 \)
### Calculation:
#### Step-by-Step Calculation of \( P(3) \):
To calculate the probability \( P(3) \), where exactly three numbers match the winning combination, we substitute the values into the hypergeometric formula.
1. **Compute the necessary factorials:**
- \( A! \)
- \( (A - x)! \)
- \( x! \)
- \( B! \)
- \( (B - n + x)! \)
- \( (n - x)! \)
- \( (A + B)! \)
- \( (A + B - n)! \)
- \( n! \)
2. **Substitute these values into the formula** and compute the result.
Finally, the output should provide the probability, rounded to four decimal places:
\[ P(3) = \boxed{} \]
**Note:** The box indicates where the calculated probability should be filled in.
### Probabilities
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