To solve this problem, we will use basic probability concepts.
(a) Find P(A)
Event A is the event that the first component drawn is defective. We know there are 210 defective components out of a total of 1082 components.
\[ P(A) = \frac{\text{Number of defective components}}{\text{Total number of components}} = \frac{210}{1082} \]
Calculating this:
\[ P(A) \approx 0.1946 \quad (\text{rounded to four decimal places}) \]
(b) Find P(B|A)
Event B is the event that the second component drawn is defective given that the first component drawn (event A) is defective. If the first component drawn is defective, one defective component has been removed, and one total component has also been removed.
So, if the first component is defective, we now have:
- Defective components remaining: 210 - 1 = 209
- Total components remaining: 1082 - 1 = 1081
Thus,
\[ P(B|A) = \frac{\text{Number of remaining defective components}}{\text{Total components remaining}} = \frac{209}{1081} \]
Calculating this:
\[ P(B|A) \approx 0.1936 \quad (\text{rounded to four decimal places}) \]
(c) Find P(A and B)
Using the definition of conditional probability, we find P(A and B) as follows:
\[ P(A \text{ and } B) = P(A) \cdot P(B|A) = P(A) \cdot P(B|A) = \frac{210}{1082} \cdot \frac{209}{1081} \]
Calculating this:
- First compute \(P(A)\):
\[ P(A) \approx 0.1946 \]
- Then compute \(P(B|A)\):
\[ P(B|A) \approx 0.1936 \]
Now, multiply these results together:
\[ P(A \text{ and } B) \approx (0.1946) \cdot (0.1936) \approx 0.0377 \quad (\text{rounded to four decimal places}) \]
(d) Are A and B independent? Is it reasonable to treat A and B as though they were independent? Explain.
Two events A and B are independent if \(P(B|A) = P(B)\). From part (a):
\[ P(B) = \frac{\text{Number of defective components}}{\text{Total number of components}} = \frac{210}{1082} \approx 0.1946 \]
From part (b):
\[ P(B|A) = \frac{209}{1081} \approx 0.1936 \]
Since \(P(B|A) \neq P(B)\), the events A and B are not independent.
Reasonableness of treating A and B as independent: It is not reasonable to treat A and B as independent because the occurrence of the first event (drawing a defective component) affects the outcome of the second event (the reduced number of defective components and total components).
In conclusion:
- (a) \(P(A) \approx 0.1946\)
- (b) \(P(B|A) \approx 0.1936\)
- (c) \(P(A \text{ and } B) \approx 0.0377\)
- (d) A and B are not independent and should not be treated as such.
1 answer