Problem 79 - A First Course in Probability

Question

Suppose that a batch of 100 items contains 6 that are defective and 94 that are not defective. If \(X\) is the number of defective items in a randomly drawn sample of 10 items from the batch, find (a) \(P\\{X=0\\}\) and \((b) P\\{X>2\\}\)

Step-by-Step Solution

Answer

The short answer to the problem is: (a) The probability of having no defective items in the sample is \(P\\{X=0\\} \approx 0.2363\). (b) The probability of having more than 2 defective items in the sample is \(P\\{X>2\\} \approx 0.6741\).

Step 1: Calculate the probability of having no defective items in the sample, \(P\\{X=0\\}\)

In this case, we need to find the probability \(P\\{X=0\\}\), that is, the probability of having no defective items in the sample. Using the hypergeometric probability formula, we have: \(P(X=0) = \frac{\binom{6}{0} \binom{94}{10}}{\binom{100}{10}}\) Now, we need to compute the individual combinations and divide.

Step 2: Evaluate the combinations in formula#for P(X=0)# - \(\binom{6}{0} = 1\) (choosing 0 defective items) - \(\binom{94}{10} = 40929782\) (choosing 10 non-defective items) - \(\binom{100}{10} = 173103094564\) (all ways of choosing 10 items from the batch of 100) Now we substitute these values into the formula: \(P(X=0) = \frac{1 \times 40929782}{173103094564}\)

Step 3: Compute the probability#for P(X=0)# \(P(X=0) = \frac{40929782}{173103094564} = 0.2363\) So, the probability of having no defective items in the sample is approximately 0.2363.

Step 4: Calculate the probability of having more than 2 defective items in the sample, \(P\\{X>2\\}\)

In this case, we need to find the probability \(P\\{X > 2\\}\), that is, the probability of having more than 2 defective items in the sample. Since there are only 6 defective items in the population and we are drawing a sample of 10 items, the possible values for the number of defective items in the sample are: 0, 1, 2, 3, 4, 5, and 6. We can calculate \(P\\{X>2\\}\) by finding the complementary probability: \(P\\{X>2\\} = 1 - P\\{X\leq 2\\} = 1 - [ P\\{X=0\\} + P\\{X=1\\} + P\\{X=2\\} ]\) Now, we need to calculate the probabilities for \(P\\{X=1\\}\) and \(P\\{X=2\\}\) using the hypergeometric probability formula from above.

Step 5: Calculate the probabilities for P(X=1) and P(X=2)

Using the same formula as in Step 1, we calculate \(P(X=1)\) and \(P(X=2)\): \(P(X=1) = \frac{\binom{6}{1} \binom{94}{9}}{\binom{100}{10}}\) \(P(X=2) = \frac{\binom{6}{2} \binom{94}{8}}{\binom{100}{10}}\) Plug in the combinations: \(P(X=1) = \frac{6 \times 1823288516}{173103094564} \approx 0.0631\) \(P(X=2) = \frac{15 \times 30667426062}{173103094564} \approx 0.0265\)

Step 6: Compute the probability for P(X>2)

Now we can substitute the calculated probabilities into our complementary probability formula from Step 4: \(P\\{X>2\\} = 1 - [ P\\{X=0\\} + P\\{X=1\\} + P\\{X=2\\} ]\) \(P\\{X>2\\} = 1 - (0.2363 + 0.0631 + 0.0265) = 1 - 0.3259 = 0.6741\) So, the probability of having more than 2 defective items in the sample is approximately 0.6741. In conclusion: (a) \(P\\{X=0\\} \approx 0.2363\) (b) \(P\\{X>2\\} \approx 0.6741\)

Key Concepts

Probability Calculation

To understand how to calculate probabilities involving defective items in a sample, it's important to grasp the concept of probability calculations involving a hypergeometric distribution. In our problem, we are interested in two specific probabilities: the probability of having no defective items and having more than two defective items in a sample of 10.

When we talk about hypergeometric distribution, unlike the binomial distribution, it does not involve replacement. This means once an item is drawn from the sample, it is not put back, affecting the outcome of successive draws. This is particularly relevant when we are dealing with a relatively small population size where what gets drawn out could significantly affect the probabilities.

Our task is to calculate probabilities like:

• Probability of getting 0 defective items in the sample (\(P\{X=0\}\)
• Probability of getting more than 2 defective items in the sample (\(P\{X>2\}\)

These probabilities help us to determine outcomes and make informed decisions based on real-world data analysis.

Combination Formula

The core of solving problems involving hypergeometric distribution lies in understanding and applying the combination formula. The combination formula is used to calculate the number of ways we can select a given number of items from a set of items, regardless of the order.

Mathematically, the combination or binomial coefficient is expressed as:\[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]where:

• \(n\) is the total number of items
• \(r\) is the number of items to choose
• \(!\) denotes factorial, which is the product of all positive integers up to that number

In our exercise, we use combinations to not only find out ways of selecting defective items but also non-defective ones. For example, for finding \(P(X=0)\), \(\binom{6}{0}\) indicates no defective items are chosen and \(\binom{94}{10}\) indicates all selected items are non-defective.

Understanding combinations is crucial in hypergeometric distribution because it involves dealing with finite, distinct, and non-replaceable samples in probability calculations.

Random Sampling

Random sampling is a statistical method used to select a subset of items from a larger population. The main idea is to ensure that the subset accurately reflects the whole population without any bias.

In this context, we are examining a batch of 100 items, from which we randomly select a sample of 10 items. We do this to assess the batch's quality, particularly checking for defective items. Random sampling is vital here to make sure every item has an equal chance of being selected, thus maintaining the integrity of the results.

Choosing randomly helps eliminate bias, meaning the sample should give an accurate snapshot of the entire population's characteristics. In probability calculations involving hypergeometric distribution, random sampling underpins the assumptions and framework for the derived results. This method is commonly used in fields such as quality control, scientific research, and surveys.

Defective Items

In many quality control scenarios, such as the one in our question, identifying defective items is crucial. These are items that fail to meet the required standard or function as intended.

In the given exercise, the batch has 6 defective items out of 100, and we are assessing probabilities related to these defects when a sample is drawn. Calculating these probabilities helps in determining the reliability and efficiency of a production process or a distribution supply.

When we refer to defective items in probability calculations, it involves understanding the impact of defects on the outcome and determining how likely it is that one or more defective items are chosen in a sample. This helps in

• Evaluating the production quality
• Predicting failure rates
• Planning improvements in manufacturing or remanufacturing processes

Assessing these defective items statistically ensures appropriate decision-making to improve the overall quality and reduce wastage.