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A foreign student club lists as its members 2 Canadians, 3 Japanese, 5 Italians, and 2 Germans. If a committee of 4 is selected at random, find the probability that (a) all nationalities are represented; (b) all nationalities except the Italians are represented.

Short Answer

Expert verified
The probability of all nationalities being represented on the committee is \(\frac{C(2, 1) \times C(3, 1) \times C(5, 1) \times C(2, 1)}{C(12, 4)}\) and the probability of all nationalities excluding the Italians being represented is \(\frac{C(2, 1) \times C(3, 1) \times C(2, 1) \times [C(2, 1) + C(3, 1) + C(2, 1)]}{C(12, 4)}\)

Step by step solution

01

Computation of Total Possible Committees

The committee can be formed by selecting 4 members out of the total number of students. This can be given by the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\), where \(n\) is the total number of objects, which is 12 in this case, and, \(k\) is the total number of items to be chosen, which is 4 in this case. Thus, the total number of possible ways to form a committee is \(C(12, 4)\).
02

Probability of all nationalities being represented

For all nationalities to be represented in a committee of 4, one member from each nationality must be chosen. So, 1 Canadian, 1 Japanese, 1 Italian, and 1 German. The number of ways this can occur is \[ C(2, 1) \times C(3, 1) \times C(5, 1) \times C(2, 1) \]The probability is then calculated by dividing the number of successful outcomes by the total possible outcomes.
03

Probability of all nationalities except the Italians being represented

For a committee of 4 to be formed excluding the Italians, one member must be chosen from the Canadians, Japanese, and Germans, leaving one spot for either additional Canadian, Japanese, or German. The number of ways this can occur is given by \[ C(2, 1) \times C(3, 1) \times C(2, 1) \times [C(2, 1) + C(3, 1) + C(2, 1)] \] The probability is then calculated by dividing the number of successful outcomes by the total possible outcomes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorial Probability
Combinatorial probability is a fundamental concept in statistics that deals with determining the likelihood of various outcomes in situations where order does not matter. It often involves counting the number of ways a certain event can occur out of all possible events. When we talk about forming committees, for example, we are looking at combinations of members, which are selections where the order isn't important.

For our foreign student club committee, the problem boils down to simple combinatorial calculations. The generic formula for combinations mentioned in the problem, \(C(n, k) = \frac{n!}{k!(n-k)!}\), is essential. Here, '!' denotes the factorial operation, which is the product of all positive integers up to the number – for instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1\). Using this combinatorial logic, we calculated the total number of possible committees.

To calculate the specific probabilities given in parts (a) and (b) we applied the same combinatorial reasoning to subsets or groups within the student club, ensuring we counted each selection of nationalities according to the specific constraints provided.
Probability of Event Outcomes
Once we have the combinatorial groundwork laid out, the next step is to apply it to find the probability of event outcomes. The probability is defined as the ratio of the number of successful outcomes over the total number of possible outcomes. In the context of our committee selection problem, successful outcomes are the number of ways we can form a committee that complies with the given criteria.

In part (a), we consider the event where all nationalities are represented. To determine the probability, we multiply the combinations of choosing one student from each nationality and divide that by the total number of ways to form a committee. Similarly, in part (b), we eliminate one nationality – the Italians – from the selection and calculate the number of ways to form a committee from the remaining students, then divide by the total number of committees.

Understanding probability calculations of event outcomes is crucial in scenarios involving risk, decision making, and predictive modeling. Clear interpretation and precise calculation of these probabilities can aid in making informed decisions based on data.
Statistics for Engineers and Scientists
Statistics for engineers and scientists encompasses a wide range of data analysis techniques and statistical theories that are critical in solving real-world problems in various technical fields. The problems in engineering and science often involve uncertainty and variability. Through statistical methods, professionals in these fields are able to make sense of the data, allowing them to model, analyze, and predict outcomes.

The probability exercise involving the foreign student club committee is a fundamental example of how statistics is applied in engineering and science. Combinatorial probability, for instance, might be used by engineers in quality control processes for determining the likelihood of defects in a batch of products. Scientists might use the same principles to analyze data from experiments or simulations.

It's important for students in these disciplines to not only know how to perform statistical calculations but also to understand the context – why these calculations are necessary and how they inform bigger decisions or lead to deeper insights in their respective fields.

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Most popular questions from this chapter

A production process produces electronic component parts. It has presumably been established that the probability of a defective part is \(0.01 .\) During a test of this presumption, 500 items are sampled randomly and 15 defective out of the 500 were observed. (a) What is your response to the presumption that the process is \(1 \%\) defective? Be sure that a computed probability accompanies your comment. (b) Under the presumption of a \(1 \%\) defective process, what is the probability that only 3 would be found defective? (c) Do (a) and (b) again using the Poisson approximation.

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During a manufacturing process 15 units are randomly selected each day from the production line to check the percent defective. From historical information it is known that the probability of a defective unit is \(0.05 .\) Any time that two or more defectives are found in the sample of \(15,\) the process is stopped. This procedure is used to provide a signal in case the probability of a defective has increased. (a) What is the probability that on any given day the production process will be stopped? (Assume \(5 \%\) defective.) (b) Suppose that the probability of a defective has increased to \(0.07 .\) What is the probability that on any given day the production process will not be stopped?

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