Problem 107 - Applied Statistics and Probability for Engineers

Question

A lot of 100 semiconductor chips contains 20 that are defective. Two are selected randomly, without replacement, from the lot (a) What is the probability that the first one selected is defective? (b) What is the probability that the second one selected is defective given that the first one was defective? (c) What is the probability that both are defective? (d) How does the answer to part (b) change if chips selected were replaced prior to the next selection?

Step-by-Step Solution

Answer

(a) \( \frac{1}{5} \); (b) \( \frac{19}{99} \); (c) \( \frac{19}{495} \); (d) \( \frac{1}{5} \).

Step 1: Calculate the probability of first chip being defective

The first chip is randomly selected from a total of 100 chips, where 20 are defective. The probability of selecting a defective chip first is calculated as the ratio of defective chips to total chips: \[ P( ext{First is defective}) = \frac{20}{100} = \frac{1}{5} \]

Step 2: Probability of second chip being defective given first is defective

If the first chip selected is defective, one defective chip is removed, leaving 19 defective out of 99 chips remaining. Therefore, the probability that the second chip is defective given the first was defective is: \[ P( ext{Second is defective} | ext{First is defective}) = \frac{19}{99} \]

Step 3: Probability that both chips are defective

To find the probability that both chips selected are defective, we use the multiplication rule since both events's probabilities are conditional. We multiply the individual probabilities from Step 1 and Step 2: \[ P( ext{Both are defective}) = P( ext{First is defective}) \times P( ext{Second is defective} | ext{First is defective}) = \frac{1}{5} \times \frac{19}{99} = \frac{19}{495} \]

Step 4: Effect of replacement on probability of second chip being defective

If the first chip is replaced before the second selection, the number of defective and total chips remain the same for each draw. Thus, the probability that the second chip is defective remains the initial probability: \[ P( ext{Second is defective with replacement}) = \frac{20}{100} = \frac{1}{5} \]

Key Concepts

Defective Chips

In the given problem, we are dealing with semiconductor chips, specifically a lot of 100 chips containing 20 defective ones. Defective chips are those that do not meet the required specifications and hence cannot be used effectively in their intended applications.
Understanding the concept of defective chips is crucial in quality control processes. When a batch contains defective chips, it's important to identify them to prevent faulty products from reaching consumers. In our exercise, knowing the portion of defective chips is key to calculating probabilities involved when selecting them randomly.
For any similar problems, always start by identifying the total number of items and how many of those are defective. This will guide the initial steps in determining the probability for picking a defective item.

Conditional Probability

Conditional probability is the probability of an event happening given that another event has already occurred. In simpler terms, it's figuring out the likelihood of something given some prior knowledge.
In the case of our chips, if we know that the first chip selected is defective, it influences or changes our calculation of getting a second defective chip. After removing one defective chip, there are now 19 defective chips left out of 99.

• Given the first chip is defective: the conditional probability becomes \( P(\text{Second is defective} | \text{First is defective}) = \frac{19}{99} \)

This is a classic example of how conditional probability can adjust our calculations when figuring out the odds under certain conditions. Understanding this concept can enhance decision-making in uncertain situations.

Multiplication Rule for Probabilities

This rule is used when we need to find the probability of two independent events happening together. However, in scenarios where one event affects the outcome of another, as in our case, the events become dependent.
When calculating the probability of both chips being defective, we observe that the probability of the second event depends on the outcome of the first. That's why:

• The formula becomes: \[ P(\text{Both are defective}) = P(\text{First is defective}) \times P(\text{Second is defective} | \text{First is defective}) \]
• Substituting the values we calculated earlier: \[ \frac{1}{5} \times \frac{19}{99} = \frac{19}{495} \]

This illustrates the multiplication rule for dependent events, highlighting the importance of acknowledging the nature of the events at play. It's essential in real-life applications where multiple events interact.