7. Suppose 99% of all dogs are good and that Weathers's fiancee/wife (depending on when you read this question) brings home 10 randomly chosen dogs. (a) What is the probability that exactly 9 of those dogs are good? Write the formula, with the appropriate values substituted in, that would give you the correct probability. There is no need to evaluate the expression for a numerical answer. (b) What is the average number of good dogs in a random sample of 10 dogs?
7. Suppose 99% of all dogs are good and that Weathers's fiancee/wife (depending on when you read this question) brings home 10 randomly chosen dogs. (a) What is the probability that exactly 9 of those dogs are good? Write the formula, with the appropriate values substituted in, that would give you the correct probability. There is no need to evaluate the expression for a numerical answer. (b) What is the average number of good dogs in a random sample of 10 dogs?
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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discrete math
![**Question 7:**
Suppose 99% of all dogs are good and that Weathers's fiancée/wife (depending on when you read this question) brings home 10 randomly chosen dogs.
(a) What is the probability that exactly 9 of those dogs are good? Write the formula, with the appropriate values substituted in, that would give you the correct probability. There is no need to evaluate the expression for a numerical answer.
(b) What is the average number of good dogs in a random sample of 10 dogs?
---
**Explanation:**
This question involves understanding probabilities in a binomial distribution scenario, where each dog can be categorized as either "good" or "not good."
(a) The probability of having exactly 9 good dogs can be calculated using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Where:
- \( n = 10 \) (total number of dogs)
- \( k = 9 \) (number of good dogs desired)
- \( p = 0.99 \) (probability of a dog being good)
Substituting these values into the formula:
\[ P(X = 9) = \binom{10}{9} (0.99)^9 (0.01)^1 \]
(b) The average number, or expected value, of good dogs in a random sample of 10 is found using the formula for expectation in a binomial distribution:
\[ E(X) = n \times p \]
Substituting the given values:
\[ E(X) = 10 \times 0.99 = 9.9 \]
This means that on average, you would expect 9.9 out of 10 dogs to be good.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7cef3050-b0aa-40af-9112-57cab04a18ba%2F1dae15a8-e81c-4a68-8565-88e0bc1ec3d2%2Fq8d0ll_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 7:**
Suppose 99% of all dogs are good and that Weathers's fiancée/wife (depending on when you read this question) brings home 10 randomly chosen dogs.
(a) What is the probability that exactly 9 of those dogs are good? Write the formula, with the appropriate values substituted in, that would give you the correct probability. There is no need to evaluate the expression for a numerical answer.
(b) What is the average number of good dogs in a random sample of 10 dogs?
---
**Explanation:**
This question involves understanding probabilities in a binomial distribution scenario, where each dog can be categorized as either "good" or "not good."
(a) The probability of having exactly 9 good dogs can be calculated using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Where:
- \( n = 10 \) (total number of dogs)
- \( k = 9 \) (number of good dogs desired)
- \( p = 0.99 \) (probability of a dog being good)
Substituting these values into the formula:
\[ P(X = 9) = \binom{10}{9} (0.99)^9 (0.01)^1 \]
(b) The average number, or expected value, of good dogs in a random sample of 10 is found using the formula for expectation in a binomial distribution:
\[ E(X) = n \times p \]
Substituting the given values:
\[ E(X) = 10 \times 0.99 = 9.9 \]
This means that on average, you would expect 9.9 out of 10 dogs to be good.
Expert Solution
Step 1
99% dogs are good :
P= 0.99
q = 1-P = 1- 0.99 = 0.01
N= 10
Binomial probability :
P(X=x) = NCx*Px *q(N-x )
Where :
NCx = N!/(x! * (N-x )!)
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