Problem 145 - Applied Statistics and Probability for Engineers

Question

A batch of 500 containers of frozen orange juice contains 5 that are defective. Two are selected, at random, without replacement, from the batch. Let \(A\) and \(B\) denote the events that the first and second containers selected are defective, respectively. (a) Are \(A\) and \(B\) independent events? (b) If the sampling were done with replacement, would \(A\) and \(B\) be independent?

Step-by-Step Solution

Answer

(a) No, not independent without replacement. (b) Yes, independent with replacement.

Step 1: Understanding the Problem

We have 500 containers, with 5 defective ones. We select two containers without replacement. Event \(A\) is choosing a defective one first, and event \(B\) is choosing a defective one second.

Step 1: Calculate Probability of Event A

There are 5 defective containers out of 500. The probability that the first container selected (event \(A\)) is defective is:\[ P(A) = \frac{5}{500} = 0.01 \]

Step 2: Calculate Probability of Event B given A

If the first container is defective, we now have 4 defective containers left out of 499 total containers. So, the probability that the second container is defective given the first one is defective (event \(B\) given \(A\)) is:\[ P(B|A) = \frac{4}{499} \]

Step 3: Calculate Probability of Event B

If we don't assume that the first container was defective, there are still 5 defective containers out of 499 containers left. So, the probability that the second container is defective (event \(B\)) is:\[ P(B) = \frac{5}{499} \]

Step 4: Determine Independence (Without Replacement)

Events \(A\) and \(B\) are independent if \(P(B|A) = P(B)\). In this case, \(P(B|A) < P(B)\) (i.e., \(\frac{4}{499} eq \frac{5}{499}\)), thus \(A\) and \(B\) are not independent.

Step 5: Independence With Replacement

If sampling is done with replacement, then after selecting the first container and putting it back, the probability for the second one remains the same:\[ P(B|A) = P(B) = \frac{5}{500} = 0.01 \]Since \(P(B|A) = P(B)\), \(A\) and \(B\) would be independent when sampling with replacement.

Key Concepts

Independent Events

In probability theory, two events are said to be independent if the occurrence of one does not affect the probability of the occurrence of the other. This means that for two events, let's call them \( A \) and \( B \), they are independent if the probability of \( A \) happening, given that \( B \) has happened \( P(A|B) \), is equal to the probability of \( A \) happening on its own \( P(A) \).
In practical terms, when you select a container without replacement, the condition of the first event affects the second. For instance, if you pick a defective container first, there are fewer defective ones left, changing the probability of the second pick.

• Without replacement, the events \( A \) and \( B \) are not independent as \( P(B|A) eq P(B) \).
• With replacement, independence is preserved because putting back the container resets the conditions for each event. In this case, \( P(B|A) = P(B) \).

Conditional Probability

Conditional Probability is the likelihood of an event occurring given that another event has already occurred. It's mathematically expressed for events \( A \) and \( B \) as \( P(B|A) \), meaning the probability of \( B \) occurring given that \( A \) has occurred.
When choosing containers without replacement, knowing the first container's state influences the chances of the second. If one container is taken, the total number of containers reduces, and if it's defective, the number of defective ones lowers too.
For example, if the first selection is defective, \( P(B|A) \) becomes \( \frac{4}{499} \) because we now have fewer defective containers to choose from and an overall smaller pool to pick. This is different from \( P(B) = \frac{5}{499} \), illustrating how \( A \)'s occurrence impacts \( B \).

• This distinction is crucial when assessing the dependency of events.
• Understanding conditional probability helps in determining if events are dependent or independent.

Combinatorial Probability

Combinatorial Probability relates to determining the likelihood of an event by analyzing the different ways it can happen, often involving combinations and permutations. This approach helps when there are multiple ways an event can occur out of a set of possibilities.
In the context of our exercise, consider the number of ways to select two defective containers out of five from the total of 500. Each choice impacts the probabilities as the group of potential selections change with each draw.
When dealing with sampling without replacement, each draw modifies the configuration of remaining options. This is expressed in combinations.

• The first draw has a probability based on the total. If it's successful (defective), the second draw adjusts to the new count \( \frac{4}{499} \).
• For independent samples, with replacement, the probability remains constant for each event, avoiding interaction between draws.

Combinatorial methods provide a structure for calculating these probabilities, especially as events become more complex.