Question

Two manufacturing companies \(M_{1}\) and \(M_{2}\) produce a certain unit that is used in an assembly plant. Company \(M_{1}\) is larger than \(M_{2}\), and it supplies the plant with twice as many units per day as \(M_{2}\) does. \(M_{1}\) also produces more defects than \(M_{2}\). Because of past experience with these suppliers, it is felt that \(10 \%\) of \(M_{1}\) 's units have some defect, whereas only \(5 \%\) of \(M_{2}\) 's units are defective. Now, suppose that a unit is selected at random from a bin in the assembly plant. (a) What is the probability that the unit was supplied by company \(M_{1} ?\) (b) What is the probability that the unit is defective? (c) What is the probability that the unit was supplied by \(M_{1}\) if the unit is defective?

Short answer

(a) \(\frac{2}{3}\); (b) \(\frac{1}{12}\); (c) \(\frac{2}{3}\).

Step-by-step solution

Define Variables and Given Information

Let the number of units supplied by \(M_2\) be \(x\). Therefore, the number of units supplied by \(M_1\) is \(2x\) because \(M_1\) supplies twice as many units as \(M_2\). The probability of selecting a unit from \(M_1\) is \(\frac{2x}{3x} = \frac{2}{3}\) and from \(M_2\) is \(\frac{x}{3x} = \frac{1}{3}\). The probability of a defect from \(M_1\) is \(0.10\) and from \(M_2\) is \(0.05\).

Calculate Probability of Unit Being from M1

The probability of a randomly selected unit being from \(M_1\) is simply the probability of picking a unit from \(M_1\), which we calculated in Step 1:\[ P(M_1) = \frac{2}{3} \]

Calculate Probability of a Defective Unit

Use the Law of Total Probability to find the probability of a defective unit:\[ P(D) = P(D|M_1) \cdot P(M_1) + P(D|M_2) \cdot P(M_2) \]\[ P(D) = 0.10 \cdot \frac{2}{3} + 0.05 \cdot \frac{1}{3} \]Calculate:\[ P(D) = \frac{2}{30} + \frac{1}{60} = \frac{5}{60} = \frac{1}{12} \]

Calculate Probability of Unit Being from M1 Given It is Defective

Use Bayes' Theorem to calculate this probability:\[ P(M_1|D) = \frac{P(D|M_1) \cdot P(M_1)}{P(D)} \]Substitute the known values:\[ P(M_1|D) = \frac{0.10 \cdot \frac{2}{3}}{\frac{1}{12}} \]Simplify:\[ P(M_1|D) = \frac{0.20}{1} = \frac{2}{3} \]

Summarize the Results

(a) The probability that the unit was supplied by \(M_1\) is \(\frac{2}{3}\).(b) The probability that the unit is defective is \(\frac{1}{12}\).(c) The probability that the unit was supplied by \(M_1\) given it is defective is \(\frac{2}{3}\).

Key concepts

Probability Calculation

Probability calculation is a fundamental concept in probability theory. It helps determine how likely an event is to occur.
In this exercise, we calculated different probabilities related to two companies supplying units.
First, we identified the probability that a unit was supplied by company \( M_1 \). Since \( M_1 \) supplies twice as many units as \( M_2 \), the probability for a unit to come from \( M_1 \) is \( \frac{2}{3} \).
Additionally, the probability for a unit to come from \( M_2 \) is \( \frac{1}{3} \). This straightforward calculation illustrates how supply rates affect probabilities.
Later in the exercise, the probability of a unit being defective is found using the Law of Total Probability, a concept we'll explore further below.

Bayes' Theorem

Bayes' Theorem is a powerful tool in probability theory for calculating conditional probabilities. It tells us how likely a hypothesis is, given new evidence. In simple terms, it reverses conditional probabilities.
In this problem, we used Bayes' Theorem to determine the probability that a unit is from company \( M_1 \) given that it is defective. The equation is:
\[ P(M_1|D) = \frac{P(D|M_1) \cdot P(M_1)}{P(D)} \]
Here's a breakdown:

• \( P(M_1|D) \) is the probability that a defective unit is from \( M_1 \).
• \( P(D|M_1) \) is the probability of a unit being defective if it is from \( M_1 \), which is \( 0.10 \).
• \( P(M_1) \) is the probability of selecting any unit from \( M_1 \), previously calculated as \( \frac{2}{3} \).
• \( P(D) \) is the total probability of a unit being defective, already calculated as \( \frac{1}{12} \).

Substitute these values into the Bayes' Theorem formula to calculate
\[ P(M_1|D) = \frac{0.10 \times \frac{2}{3}}{\frac{1}{12}} = \frac{0.20}{1} = \frac{2}{3} \].
This result gives clearer insight into how likely it is that a defective unit is from \( M_1 \) compared to \( M_2 \). Bayes' Theorem shows how new data, like defect status, changes prior outcomes.

Law of Total Probability

The Law of Total Probability helps calculate the probability of an event based on partitions of the event space. It effectively uses known probabilities to find the total probability of complex events.
In the exercise, we needed to find the probability of a unit being defective. This probability had to account for whether units were from \( M_1 \) or \( M_2 \).
The formula used was:
\[ P(D) = P(D|M_1) \cdot P(M_1) + P(D|M_2) \cdot P(M_2) \]
Here's the breakdown:

• \( P(D|M_1) \) is the probability of a defect from \( M_1 \), which is \( 0.10 \).
• \( P(M_1) \) is the prior probability of a unit coming from \( M_1 \), calculated as \( \frac{2}{3} \).
• \( P(D|M_2) \) is the probability of a defect from \( M_2 \), \( 0.05 \).
• \( P(M_2) \) is the prior probability of a unit coming from \( M_2 \), \( \frac{1}{3} \).

Upon solving, we find:
\[ P(D) = 0.10 \cdot \frac{2}{3} + 0.05 \cdot \frac{1}{3} = \frac{1}{12} \]
The Law of Total Probability thus synthesizes known information to derive the probability of a defective unit, factoring in contributions from different suppliers.