Refer to these values from the previous step. N = 13, M = 7, n = 6, k = 5, NM = 6, n - k = 1 Substitute these values into the hypergeometric probability formula, or use technology. Round your answer to four decimal places. CMC-k N-M P(x = k) = P(x = 5) = c57C₁6 13 N Therefore, the probability of selecting five brown and one red M&Ms from a candy dish containing seven brown and six red M&Ms is
Refer to these values from the previous step. N = 13, M = 7, n = 6, k = 5, NM = 6, n - k = 1 Substitute these values into the hypergeometric probability formula, or use technology. Round your answer to four decimal places. CMC-k N-M P(x = k) = P(x = 5) = c57C₁6 13 N Therefore, the probability of selecting five brown and one red M&Ms from a candy dish containing seven brown and six red M&Ms is
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
Step 5
![Step 4
Using the values we have previously determined, \( N = 13 \), \( M = 7 \), \( n = 6 \), and \( k = 5 \). The remaining values needed for the formula are \( N - M \) and \( n - k \).
Find \( N - M \).
\[ N - M = 13 - 7 \]
\[ = 6 \]
Find \( n - k \).
\[ n - k = 6 - 5 \]
\[ = 1 \]
---
Step 5
Refer to these values from the previous step.
\[ N = 13, M = 7, n = 6, k = 5, N - M = 6, n - k = 1 \]
Substitute these values into the hypergeometric probability formula, or use technology. Round your answer to four decimal places.
\[ P(x = k) = \frac{\binom{M}{k} \binom{N-M}{n-k}}{\binom{N}{n}} \]
\[ P(x = 5) = \frac{\binom{7}{5} \binom{6}{1}}{\binom{13}{6}} \]
\[ = \]
Therefore, the probability of selecting five brown and one red M&Ms from a candy dish containing seven brown and six red M&Ms is \(\boxed{}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff0fdd9be-9990-44a9-8005-60989cbb2f12%2F5d2e925e-aefd-4639-b1e1-ad3221bee05f%2Fapg5kea_processed.png&w=3840&q=75)
Transcribed Image Text:Step 4
Using the values we have previously determined, \( N = 13 \), \( M = 7 \), \( n = 6 \), and \( k = 5 \). The remaining values needed for the formula are \( N - M \) and \( n - k \).
Find \( N - M \).
\[ N - M = 13 - 7 \]
\[ = 6 \]
Find \( n - k \).
\[ n - k = 6 - 5 \]
\[ = 1 \]
---
Step 5
Refer to these values from the previous step.
\[ N = 13, M = 7, n = 6, k = 5, N - M = 6, n - k = 1 \]
Substitute these values into the hypergeometric probability formula, or use technology. Round your answer to four decimal places.
\[ P(x = k) = \frac{\binom{M}{k} \binom{N-M}{n-k}}{\binom{N}{n}} \]
\[ P(x = 5) = \frac{\binom{7}{5} \binom{6}{1}}{\binom{13}{6}} \]
\[ = \]
Therefore, the probability of selecting five brown and one red M&Ms from a candy dish containing seven brown and six red M&Ms is \(\boxed{}\).
![**Tutorial Exercise: Hypergeometric Distribution**
**Scenario:**
A candy dish contains seven brown and six red M&Ms. A child selects six M&Ms without checking the colors. What is the probability that there are five brown and one red M&M in the selection?
**Step 1: Determine the Appropriateness of the Hypergeometric Distribution**
- Begin by confirming that the hypergeometric distribution is appropriate for this sample.
- The proportion, \( \frac{n}{N} = \frac{6}{13} \). Calculate this value:
\[
\frac{n}{N} = 0.46
\]
- Check if \( \frac{n}{N} \geq 0.05 \), where \( n = 6 \) (the number of M&Ms being selected), and \( N = 13 \) (the total number of M&Ms).
- Since 0.46 is greater than 0.05, we can use the hypergeometric probability distribution.
**Step 2: Define Variables for the Hypergeometric Distribution**
- Recall that for a hypergeometric distribution, the population contains \( M \) successes and \( N - M \) failures.
- Brown M&M is considered a success, red M&M a failure.
- Given seven brown and six red M&Ms, the total number of M&Ms is \( N = 13 \) and successes (brown M&Ms) \( M = 7 \).
**Step 3: Calculate the Probability**
- Use the formula for the probability of exactly \( k \) successes in a sample of size \( n \):
\[
P(x = k) = \frac{C_k^M C_{n-k}^{N-M}}{C_n^N}
\]
- With seven brown M&Ms in a bowl of thirteen, \( M = 7 \), \( N = 13 \).
- For a sample size \( n = 6 \):
- \( k = 5 \) (number of successes or brown M&Ms).
- We are calculating the probability of exactly five brown M&Ms when six are drawn.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff0fdd9be-9990-44a9-8005-60989cbb2f12%2F5d2e925e-aefd-4639-b1e1-ad3221bee05f%2Fxp3qp1_processed.png&w=3840&q=75)
Transcribed Image Text:**Tutorial Exercise: Hypergeometric Distribution**
**Scenario:**
A candy dish contains seven brown and six red M&Ms. A child selects six M&Ms without checking the colors. What is the probability that there are five brown and one red M&M in the selection?
**Step 1: Determine the Appropriateness of the Hypergeometric Distribution**
- Begin by confirming that the hypergeometric distribution is appropriate for this sample.
- The proportion, \( \frac{n}{N} = \frac{6}{13} \). Calculate this value:
\[
\frac{n}{N} = 0.46
\]
- Check if \( \frac{n}{N} \geq 0.05 \), where \( n = 6 \) (the number of M&Ms being selected), and \( N = 13 \) (the total number of M&Ms).
- Since 0.46 is greater than 0.05, we can use the hypergeometric probability distribution.
**Step 2: Define Variables for the Hypergeometric Distribution**
- Recall that for a hypergeometric distribution, the population contains \( M \) successes and \( N - M \) failures.
- Brown M&M is considered a success, red M&M a failure.
- Given seven brown and six red M&Ms, the total number of M&Ms is \( N = 13 \) and successes (brown M&Ms) \( M = 7 \).
**Step 3: Calculate the Probability**
- Use the formula for the probability of exactly \( k \) successes in a sample of size \( n \):
\[
P(x = k) = \frac{C_k^M C_{n-k}^{N-M}}{C_n^N}
\]
- With seven brown M&Ms in a bowl of thirteen, \( M = 7 \), \( N = 13 \).
- For a sample size \( n = 6 \):
- \( k = 5 \) (number of successes or brown M&Ms).
- We are calculating the probability of exactly five brown M&Ms when six are drawn.
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