Problem 47 - Applied Statistics and Probability for Engineers

Question

In a chemical plant, 24 holding tanks are used for final product storage. Four tanks are selected at random and without replacement. Suppose that six of the tanks contain material in which the viscosity exceeds the customer requirements. (a) What is the probability that exactly one tank in the sample contains high- viscosity material? (b) What is the probability that at least one tank in the sample contains high-viscosity material? (c) In addition to the six tanks with high viscosity levels, four different tanks contain material with high impurities. What is the probability that exactly one tank in the sample contains high-viscosity material and exactly one tank in the sample contains material with high impurities?

Step-by-Step Solution

Answer

(a) 0.4607, (b) 0.712, (c) 0.2055.

Step 1: Determine the Probability of Exactly One High-Viscosity Tank

We need to find the probability that in a sample of 4 tanks, exactly 1 has high viscosity. We'll use the hypergeometric distribution for this calculation.The hypergeometric distribution probability mass function is given by:\[ P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}} \]where:- \( N = 24 \) is the total number of tanks.- \( K = 6 \) is the number of high-viscosity tanks.- \( n = 4 \) is the number of tanks in the sample.- \( k = 1 \) is the number of high-viscosity tanks in the sample.Substituting the values:\[ P(1) = \frac{{\binom{6}{1} \cdot \binom{18}{3}}}{{\binom{24}{4}}} \]Calculating each part:- \( \binom{6}{1} = 6 \)- \( \binom{18}{3} = 816 \)- \( \binom{24}{4} = 10,626 \)The probability is:\[ P(1) = \frac{6 \times 816}{10,626} = \frac{4896}{10,626} \approx 0.4607 \]

Step 2: Probability of At Least One High-Viscosity Tank

To find the probability that at least one tank in the sample contains high-viscosity material, we use the complement rule:\[ P(\text{At least 1}) = 1 - P(\text{None}) \]First, calculate the probability of none of the 4 tanks having high viscosity using the hypergeometric distribution:\[ P(\text{None}) = \frac{{\binom{6}{0} \cdot \binom{18}{4}}}{{\binom{24}{4}}} \]Calculate each part:- \( \binom{6}{0} = 1 \)- \( \binom{18}{4} = 3,060 \)- \( \binom{24}{4} = 10,626 \)The probability is:\[ P(\text{None}) = \frac{1 \times 3,060}{10,626} = \frac{3,060}{10,626} \approx 0.288 \]Therefore:\[ P(\text{At least 1}) = 1 - 0.288 = 0.712 \]

Step 3: Determine the Probability of One High-Viscosity and One High-Impurity Tank

We need to find the probability that exactly one tank in the sample contains high-viscosity material and exactly one contains high-impurity material. First, calculate the number of favorable ways to choose 1 high-viscosity and 1 high-impurity tank. Then, calculate how to fill the remaining 2 tanks out of normal tanks.- Number of high-viscosity tanks \( K_h = 6 \)- Number of high-impurity tanks \( K_i = 4 \)- Number of tanks that are neither \( N - K_h - K_i = 24 - 6 - 4 = 14 \)Use combinations:\[ \binom{6}{1} \times \binom{4}{1} \times \binom{14}{2} \]Calculate each part:- \( \binom{6}{1} = 6 \)- \( \binom{4}{1} = 4 \)- \( \binom{14}{2} = 91 \)Now, calculate the total number of ways to choose 4 tanks out of 24:\[ \binom{24}{4} = 10,626 \]Finally, calculate the probability:\[ P(1 \, \text{high-viscosity and 1 \, high-impurity}) = \frac{6 \times 4 \times 91}{10,626} = \frac{2,184}{10,626} \approx 0.2055 \]

Key Concepts

Probability Calculation

In probability theory, the calculation of probability can help you estimate the likelihood of an event occurring. In scenarios involving a finite population, a hypergeometric distribution often comes in handy. This distribution is ideal when you draw objects from a group without replacing them, as in our case with the holding tanks. For a probability calculation using the hypergeometric distribution, the main focus is on the probability mass function (PMF). The PMF is defined as:

• \[P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}}\]
• Where:
  • \(N\) is the total number of objects.
  • \(K\) is the number of successful outcomes in the population.
  • \(n\) is the number of draws.
  • \(k\) is the number of successful outcomes you are interested in among the draws.

This formula takes into account all possible ways to achieve the desired outcome over all possible outcomes. It is particularly useful in calculating probabilities in sampling without replacement, which is the basis for hypergeometric distribution.

Combinatorics

Combinatorics is the mathematical study of counting, arrangement, and combination of objects. It plays a significant role in probability calculations because it allows us to determine the number of possible outcomes. The main tool in combinatorics that is used for hypergeometric distribution is the binomial coefficient, often denoted as \( \binom{n}{k} \). This represents the number of ways to choose \(k\) objects from \(n\) objects without regard to the order of selection.

• For example, in the hypergeometric probability calculations, \( \binom{6}{1} \) allows us to count how many ways there are to pick 1 successful object (high-viscosity tank) from 6 possible successful objects.
• Likewise, \( \binom{18}{3} \) indicates the number of ways to select 3 from the 18 non-successful objects in the population.
• The use of \( \binom{24}{4} \) shows us how to identify the number of possible ways to choose any 4 from a total of 24 tanks.

Through combinatorics, we can quantify objective probabilities by determining the number of favorable outcomes compared to all possible outcomes, which is essential for hypergeometric distributions.

Sampling Without Replacement

Sampling without replacement is a key principle in probability and statistics. It refers to a scenario where each selected sample is not returned to the population before the next selection. This affects the probability because the population size is reduced with each draw, altering the sample space. Unlike sampling with replacement, where each draw is independent, sampling without replacement involves dependent samples. In the case of the chemical plant problem, when tanks are selected randomly for inspection, those picked are not replaced. This impacts future selections as the probabilities change dynamically with each draw.

• Firstly, when a high-viscosity tank is chosen, it cannot be picked again, reducing both the total number of tanks and the number of high-viscosity tanks available for selection.
• Consequently, subsequent selections are influenced by the previous choices, making the calculations more complex than if the sampling were done with replacement.

This dependency must be accurately modeled in probability calculations, which is why the hypergeometric distribution is imperative, as it precisely accounts for these changes in population dynamics and provides the structure needed to solve problems involving dependent samples.