Problem 7 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

A shop receives a batch of 1000 cheap lamps. The odds that a lamp is defective are \(0.1 \%\). Let \(X\) be the number of defective lamps in the batch. a. What kind of distribution does \(X\) have? What is/are the value(s) of parameter(s) of this distribution? b. What is the probability that the batch contains no defective lamps? One defective lamp? More than two defective ones?

Step-by-Step Solution

Answer

X follows a Binomial distribution with parameters n=1000 and p=0.001. P(X=0)≈0.3679, P(X=1)≈0.367, P(X>2)≈0.0811.

Step 1: Identify the type of distribution

Since the problem involves a fixed number of trials (1000 lamps) with each trial having two possible outcomes (defective or not defective), and a constant probability of being defective for each lamp, the random variable \(X\) follows a Binomial distribution.

Step 2: Determine the parameters of the distribution

The Binomial distribution is characterized by two parameters: the number of trials \(n\) and the probability of success \(p\). Here, \(n = 1000\) and \(p = 0.1\% = 0.001\). So, \(X \sim \text{Binomial}(1000, 0.001)\).

Step 3: Probability of zero defective lamps

To find the probability that there are no defective lamps \(P(X=0)\):\[P(X = 0) = \binom{1000}{0} (0.001)^0 (0.999)^{1000} = (0.999)^{1000}\]This calculates as \(P(X = 0) \approx e^{-1} \approx 0.3679\).

Step 4: Probability of one defective lamp

To find the probability of exactly one defective lamp \(P(X=1)\):\[P(X = 1) = \binom{1000}{1} (0.001)^1 (0.999)^{999}\]This calculates as \(P(X = 1) \approx 0.367\).

Step 5: Probability of more than two defective lamps

To find the probability that there are more than two defective lamps \(P(X>2)\):\[P(X>2) = 1 - P(X=0) - P(X=1) - P(X=2)\]Calculate \(P(X=2)\):\[P(X = 2) = \binom{1000}{2} (0.001)^2 (0.999)^{998}\]This approximates to \(P(X = 2) \approx 0.184\).Thus, \(P(X>2) = 1 - 0.3679 - 0.367 - 0.184 \approx 0.0811\).

Key Concepts

Probability Calculation

Probability calculation is an essential part of understanding how likely an event is to occur. In the context of this exercise, we are dealing with the probability of finding defective lamps in a batch.
The calculation here uses the Binomial distribution, which helps us determine the likelihood of a certain number of successes (in this case, defective lamps) in a fixed number of trials. Think of it like flipping a coin multiple times and counting the number of heads.
The probability of having zero defective lamps is calculated as the product of the probability of a single lamp being non-defective raised to the power of 1000 (as there are 1000 lamps) using the formula:
\[P(X = 0) = (0.999)^{1000}\]
Similarly, to find the probability of exactly one defective lamp, we multiply the probability of one lamp being defective with that of the remaining 999 lamps being non-defective:
\[P(X = 1) = 1000 \times 0.001 \times (0.999)^{999}\]
Each of these numbers tells us how likely a certain outcome is based on our fixed parameters.

Defective Items

When discussing defective items, we are referring to products that do not meet expected quality standards.
In this exercise, the defective lamps represent the occasional errors in manufacturing or handling that lead to a product not functioning as intended.
Understanding the defect rate, which is 0.1% for each lamp, allows us to use statistical models to predict how often defects might occur in larger batches.

• For small percentages like 0.1%, most batches will have few or no defects.
• Occasionally, you may find a batch with more than expected due to the random nature of occurrences.

By knowing the defect rate, producers can improve their quality control processes and consumers can assess risks. This predictability is crucial in manufacturing, where consistency matters.

Discrete Distribution

Discrete distributions are about outcomes that are countable and distinct. The Binomial distribution used here is one such example, where we're counting the number of defective items out of a total number of trials (in this case, lamps).
With discrete distributions, each possible outcome has a certain probability associated with it, which adds up to 1 when considering all possible outcomes.
In a binomial scenario:

• The number of trials, denoted as \(n\), is a fixed number. Here, \(n = 1000\).
• The probability of success, \(p\), is the chance of one trial resulting in a defect, \(p = 0.001\) for each lamp.

By knowing these parameters, calculating the probabilities of 0, 1, or more defective lamps consists of using formulas to combine individual probabilities. Discrete distributions enable us to predict and quantify these outcomes in a structured way.