Question

A certain manufactured item is visually inspected by two different inspectors. When a defective item comes through the line, the probability that it gets by the first inspector is \(.1 .\) Of those that get past the first inspector, the second inspector will "miss" 5 out of \(10 .\) What fraction of the defective items get by both inspectors?

Short answer

Answer: The probability of a defective item getting past both inspectors is 1/20.

Step-by-step solution

Probability of the first inspector missing a defective item

The probability that the first inspector misses a defective item is given as 0.1.

Probability of the second inspector missing a defective item given it passed the first inspector

The probability that the second inspector misses a defective item, given it has already passed the first inspector, is 5 out of 10, which can be simplified as 0.5.

Calculate the probability of both inspectors missing a defective item

To find the probability of the defective item getting past both inspectors, we need to multiply the probabilities from step 1 and step 2: Probability = (Probability of first inspector missing) * (Probability of second inspector missing | passed first inspector) Probability = 0.1 * 0.5 = 0.05

Express the result as a fraction

The probability of both inspectors missing a defective item is 0.05. To express this as a fraction, we can convert it as follows: 0.05 = 5/100 = 1/20 So, the fraction of defective items that get by both inspectors is 1/20.

Key concepts

Conditional Probability

Conditional probability is the chance that an event will occur given that another event has already occurred. This concept is crucial in understanding complex probability scenarios. In our exercise, we're examining a two-stage inspection process of defective items. Specifically, we want to know the probability of a defective item not being caught by the second inspector if it already managed to slip by the first one. The provided solution breaks this down by first establishing the initial probability of the first inspector missing the defect (.1), and then giving the conditional probability for the second inspector, which is 0.5. This means that the second event (second inspector missing the defect) depends on the outcome of the first event (first inspector missing the defect).

One way to think of conditional probability is with the formula:
\( P(A|B) = \frac{P(A \cap B)}{P(B)} \)
where \( P(A|B) \) is the probability of event A occurring given that B has occurred. But in cases where the events are independent, like in our case, we can simply multiply the probabilities of the two events to get the conditional probability of both occurring. It's important for students to understand that conditional probability is not always intuitive; it requires careful consideration of how one event affects another.

Independent Events

In probability theory, events are termed independent if the occurrence of one does not affect the probability of the other occurring. This principle allows us to perform simple probability calculations, such as multiplying the chance of two events happening when the events are independent. In our textbook exercise, the inspection processes are essentially independent events. The second inspector's ability to spot defects is not influenced by whether the first inspector has already missed the defect.

To strengthen this understanding, consider the formula:
\( P(A \text{ and } B) = P(A) \times P(B) \)
if and only if events A and B are independent. Despite their independence, it's crucial to note that our interpretation of the inspectors' checks assumes independence based on the information given. In real-world scenarios, independence is not always a given and must be determined through analysis and understanding of the context.

Probability Calculation

Probability calculation is the process of determining the likelihood of an event’s occurrence. It is the fundamental concept underlying all probability problems. In the context of our inspection problem, we are tasked with a straightforward probability calculation: multiplying the independent probabilities of two inspections. By presenting the solution in a step-by-step manner, the exercise shows students how to break down a problem into smaller parts, calculate each probability, and then combine them to find the total probability of a sequence of events.

Using the formula for independent events previously discussed, we compute:
\( 0.1 \times 0.5 = 0.05 \)
This calculation yields a decimal probability of 0.05, or 5%, which is equivalent to the fraction \( \frac{1}{20} \). Being able to convert between decimals, percentages, and fractions is also a valuable skill in probability calculations. Providing this step-by-step approach in exercises helps students build confidence in their ability to tackle probability problems and reinforces their computational skills.