To solve the problem, let's first summarize the information provided:
- Total components = 11
- Defective components = 4
- Non-defective components = 11 - 4 = 7
We'll compute the probabilities step by step.
A) Find \( P(A) \)
The probability of event \( A \) (the first component drawn is defective) is given by the ratio of defective components to the total components.
\[ P(A) = \frac{\text{Number of defective components}}{\text{Total number of components}} = \frac{4}{11} \]
B) Find \( P(B|A) \)
To find \( P(B|A) \), we need the probability that the second component drawn is defective, given that the first component drawn was defective. If the first component is defective, there are now:
- Defective components left = 4 - 1 = 3
- Total components left = 11 - 1 = 10
So, the probability that the second component drawn is defective, given that the first one was defective is:
\[ P(B|A) = \frac{\text{Number of defective components left}}{\text{Total number of components left}} = \frac{3}{10} \]
C) Find \( P(A \text{ and } B) \)
The joint probability \( P(A \text{ and } B) \) can be calculated as follows:
\[ P(A \text{ and } B) = P(A) \times P(B|A) = \frac{4}{11} \times \frac{3}{10} \]
Calculating this gives:
\[ P(A \text{ and } B) = \frac{4 \cdot 3}{11 \cdot 10} = \frac{12}{110} = \frac{6}{55} \]
Converting this to a decimal for formatting:
\[ P(A \text{ and } B) \approx 0.1091 \text{ (rounded to four decimal places)} \]
D) Are \( A \) and \( B \) independent? Explain.
Two events \( A \) and \( B \) are independent if and only if:
\[ P(A \text{ and } B) = P(A) \times P(B) \]
To check independence, we need to calculate \( P(B) \), the probability that the second component is defective. Without conditioning on the first component, consider the total amount.
The total possible configurations for the second component drawn can be broken down into two cases based on whether the first component was defective or not.
- If the first is defective: There are 3 defective options and 10 remaining components.
\[ P(B | A) = \frac{3}{10} \]
- If the first is non-defective: There remain 4 defective options among 10 remaining components.
\[ P(B | A^c) = \frac{4}{10} = \frac{2}{5} \]
Since we need the total probability of \( B \):
\[ P(B) = P(A) \cdot P(B|A) + P(A^c) \cdot P(B|A^c) \]
We already found \( P(B|A) \) and \( P(A) \):
\[ P(A^c) = 1 - P(A) = 1 - \frac{4}{11} = \frac{7}{11} \]
Now substituting these values:
\[ P(B) = \frac{4}{11} \times \frac{3}{10} + \frac{7}{11} \times \frac{2}{5} \]
Calculating each part:
- \(\frac{4}{11} \times \frac{3}{10} = \frac{12}{110}\)
- \(\frac{7}{11} \times \frac{2}{5} = \frac{14}{55} = \frac{28}{110}\)
Adding these two:
\[ P(B) = \frac{12}{110} + \frac{28}{110} = \frac{40}{110} = \frac{4}{11} \]
Now we found \( P(A) \) and \( P(B) \), so check independence:
\[ P(A) \times P(B) = \left(\frac{4}{11}\right) \times \left(\frac{4}{11}\right) = \frac{16}{121} \neq P(A \text{ and } B) = \frac{6}{55} \approx 0.1091 \]
Thus, \( P(A \text{ and } B) \) does not equal \( P(A) \times P(B) \). Therefore, events \( A \) and \( B \) are not independent.
Final Summary:
A. \( P(A) = \frac{4}{11} \)
B. \( P(B|A) = \frac{3}{10} \)
C. \( P(A \text{ and } B) = \frac{6}{55} \approx 0.1091 \)
D. Events \( A \) and \( B \) are not independent.
1 answer