- The probability of having exactly \(2\) defective parts is given directly by the table: \(P(X=2) = \boxed{0.04}\).
- The probability of finding at least one defect is calculated as the complement of finding zero defects: \(P(X \ge 1) = 1 - P(X=0) = 1 - 0.85 = \boxed{0.15}\).
Explanation
- Analyze the probability distribution of defective parts
The problem presents a probability distribution for the number of defective parts, denoted by the random variable \(X\). The possible outcomes for \(X\) are \(\{0, 1, 2, 3\}\), and their corresponding probabilities are given as follows:
- \(P(X=0) = 0.85\)
- \(P(X=1) = 0.10\)
- \(P(X=2) = 0.04\)
- \(P(X=3) = 0.01\)
To ensure the distribution is valid, we check that the sum of the probabilities is \(0.85 + 0.10 + 0.04 + 0.01 = 1.00\). Since the sum is 1, the distribution is valid and complete. Our goal is to determine the probability of finding exactly two defective parts in a day and the probability of finding at least one defective part.
- Calculate the probability of exactly 2 defective parts
To find the probability of having exactly \(2\) defective parts, we look directly at the provided probability distribution table for the event \(X=2\). According to the data, \(P(X=2) = 0.04\). This means there is a \(4\%\) chance that Shelia will find exactly two defective parts on any given day.
- Calculate the probability of finding any defects
To find the probability that Shelia finds at least one defect, we are looking for the probability of the event \(X \ge 1\). This event includes the cases where \(X=1\), \(X=2\), or \(X=3\). We can calculate this in two ways.
Method 1: Sum the individual probabilities for \(X=1\), \(X=2\), and \(X=3\):
\[P(X \ge 1) = P(X=1) + P(X=2) + P(X=3) = 0.10 + 0.04 + 0.01 = 0.15\]
Method 2: Use the complement rule, which states that the probability of at least one defect is equal to \(1\) minus the probability of having zero defects:
\[P(X \ge 1) = 1 - P(X=0) = 1 - 0.85 = 0.15\]
Both methods yield the same result, confirming that there is a \(15\%\) chance that Shelia will find at least one defective part on a given day.
Examples
Probability distributions like this are essential in manufacturing quality control. By understanding the likelihood of various numbers of defects, a company can plan its maintenance schedules, estimate waste, and set realistic production targets. For example, if the probability of having 3 defects is too high, the company might decide to recalibrate the machinery. Calculation: If a company produces 100 parts, the expected number of defects is \(E[X] = \sum (x \cdot P(x)) = (0 \cdot 0.85) + (1 \cdot 0.10) + (2 \cdot 0.04) + (3 \cdot 0.01) = 0.10 + 0.08 + 0.03 = 0.21\). This informs the factory manager that, on average, they will see 0.21 defects per day.