Problem 28 - A First Course in Probability

Question

A sample of 3 items is selected at random from a box containing 20 items of which 4 are defective. Find the expected number of defective items in the sample.

Step-by-Step Solution

Answer

The expected number of defective items in the sample of 3 items selected at random from a box containing 20 items, out of which 4 are defective, is approximately 0.633.

Step 1: Calculate Total Outcomes

First find the total number of ways to choose a sample of 3 items from a group of 20 items. We'll use combinations for this: \(\text{Total combinations} = {20\choose3}=\frac{20!}{3! \times (20-3)!}=1140 \) So, there are a total of 1140 ways to choose a sample of 3 items from the box.

Step 2: Calculate Combinations of Defective and Non-defective Items

Next, we will find the number of ways to choose different combinations of defective and non-defective items: 1. All non-defective items (0 defective, 3 non-defective) - \({16\choose3}\) 2. One defective and two non-defective items - \({4\choose1}\times {16\choose2}\) 3. Two defective and one non-defective item - \({4\choose2}\times {16\choose1}\) 4. All three defective items - \({4\choose3}\)

Step 3: Calculate Probabilities of Each Combination

Now we calculate the probability of each combination by dividing the number of ways to choose that combination by the total outcomes: 1. All non-defective items: \(\frac{{16\choose3}}{1140}\) 2. One defective and two non-defective items: \(\frac{{4\choose1}\times {16\choose2}}{1140}\) 3. Two defective and one non-defective item: \(\frac{{4\choose2}\times {16\choose1}}{1140}\) 4. All three defective items: \(\frac{{4\choose3}}{1140}\)

Step 4: Calculate the Expected Number of Defective Items

Now we will find the expected number of defective items by multiplying the number of defective items in each case by the probability of that case and adding those products together: Expected number of defective items = (0 defective items) * (probability of 0 defective) + (1 defective item) * (probability of 1 defective) + (2 defective items) * (probability of 2 defective) + (3 defective items) * (probability of 3 defective) Expected number of defective items = \(0\times\frac{{16\choose3}}{1140} + 1\times\frac{{4\choose1}\times {16\choose2}}{1140} + 2\times\frac{{4\choose2}\times {16\choose1}}{1140} + 3\times\frac{{4\choose3}}{1140} \) After calculating the probabilities for each case, we get: Expected number of defective items = \(0\times0.439 + 1\times0.495 + 2\times0.060 + 3\times0.006 = 0 + 0.495 + 0.120 + 0.018 = 0.633\) Therefore, the expected number of defective items in the sample is approximately 0.633.

Key Concepts

Expected Value

Understanding the concept of expected value is crucial when predicting outcomes in probability and statistics. It represents the average outcome if an experiment or process is repeated a large number of times, considering all possible occurrences and their corresponding probabilities.

For a discrete random variable, the expected value is calculated by multiplying each possible outcome by the probability of that outcome and then adding all these products together. Using the given problem as an example:

• Determine all possible outcomes and their probabilities.
• Calculate the product of each outcome with its probability.
• Sum these products to obtain the expected value.

Applying this to our scenario, if we repeatedly take samples of three items from the box, on average, we'd expect approximately 0.633 defective items per sample. Remember, while the expected value can provide insight, it doesn't guarantee results for a single trial, as it considers the long-term average.

Combinatorics

Combinatorics is the branch of mathematics dealing with combinations, permutations, and the counting of potential outcomes in a given set. It is essential in calculating probabilities as it helps us determine the number of ways events can occur.

In probability problems, we often use the combination formula — denoted as \( {n\choose k} \) — which calculates how many ways we can choose \( k \) items from a larger set of \( n \) items without considering the order. This formula is expressed as \( {n\choose k} = \frac{n!}{k!(n - k)!} \).

Applying Combinatorics to the Problem

For the textbook exercise, we were looking for the number of ways to select 3 items (\( k \)) from 20 (\( n \)), disregarding the sequence in which they're picked. This total affects the overall probability of each possible outcome involving defective items and is the foundation for understanding probability calculations in such scenarios.

Probability Calculations

Probability calculations involve determining the likelihood of various outcomes. They are expressed as a value between 0 (impossible event) and 1 (certain event). To calculate the probability of an event, we divide the number of favorable outcomes by the total number of possible outcomes.

In our exercise, we calculated different event probabilities such as having zero, one, two, or three defective items in our sample of 3 from the total of 20. It's a two-step procedure:

• First, count the number of ways each event can occur using combinatorics.
• Then, divide this by the total number of outcomes to get the probability of each event.

When combined with the principles of expected value, these probabilities enable us to forecast outcomes and make informed decisions based on the likelihood of various results.