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Find the probabilities. \(\frac{C_{1}^{2} C_{1}^{1}}{C_{2}^{3}}\)

Short Answer

Expert verified
Based on the given expression \(\frac{C_{1}^{2} C_{1}^{1}}{C_{2}^{3}}\), it is not possible to find the probability as it contains an undefined term. Please check the problem statement and make sure all numbers in the expression are entered correctly.

Step by step solution

01

Understand the combinations formula

The combinations formula calculates the number of ways to choose \(r\) items from a set of \(n\) items without considering the order. The formula for combinations is given by \(C_n^r = \frac{n!}{r!(n-r)!}\), where \(n\) is the total number of items, \(r\) is the number of items taken at a time, and \(!\) denotes a factorial.
02

Calculate combinations in the numerator

The given expression is \(\frac{C_{1}^2 C_{1}^1}{C_{2}^3}\). We first need to calculate the combinations in the numerator: \(C_{1}^{2} = \frac{1!}{2!(1-2)!} = \frac{1}{2!(-1)!}\) - (Note: By definition, \((-1)!\) is undefined.) Since \((-1)!\) is undefined, the expression becomes invalid, and it is not possible to find the probability. If you meant to write a valid expression, please check the problem statement and make sure all numbers in the expression are entered correctly.

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Key Concepts

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

Understanding Factorials
Factorials are a fundamental concept in combinatorics, the branch of mathematics that deals with counting, both as distinct, ordered arrangements (permutations), and as selection of subsets (combinations). The factorial of a non-negative integer, denoted by an exclamation point (!), represents the product of all positive integers less than or equal to the number. For example, the factorial of four (4!) is calculated as
4 = 4 3 2 1 = 24.

It is essential to note that the factorial of zero (0!) is defined to be 1 by convention. Factorials increase very rapidly with the growth of the number, and thus are extraordinarily significant while calculating combinations and understanding probability spaces in larger sets. In probability and statistics, factorials are utilized to find the number of ways objects can be arranged, and they set the groundwork for calculating probabilities in discrete sample spaces.
The Combinations Formula
The combinations formula helps determine the number of ways a subset of items can be chosen from a larger set, where the order doesn't matter. This concept is crucial when dealing with problems where you are interested in groups of items, but not in the sequence in which they appear. For instance, when you are picking team members, it does not necessarily matter who was picked first or last, but rather who is on the team.

The formula for finding the number of combinations is as follows:
C_n^r = frac{n!}{r!(n-r)!},

where n is the total number of items, and r is the number of items to choose. The formula is a fraction with the factorial of the total number of items at the numerator and the product of the factorial of the number of items chosen and the factorial of the difference between the total and the chosen items as the denominator. One critical point to remember is that the value of r cannot be greater than n, and both should be non-negative integers; this assures that the combination is a valid one.
Probability Calculation Basics
Probability calculation plays a pivotal role in determining the likelihood of different outcomes in uncertain situations. It is a measure ranging from 0 to 1, where 0 indicates impossibility, and 1 indicates certainty. In the context of combinations, the probability of a certain combination occurring is calculated by dividing the number of favorable outcomes (the specific combination) by the total number of possible outcomes. Importantly, all outcomes must be equally likely for this basic probability calculation to be valid.

For example, when you flip a coin, there are two possible outcomes: heads or tails. The probability of getting heads is frac{1}{2}, because there is one favorable outcome (getting heads) and two possible outcomes in total.

In the textbook problem you mentioned, the probability calculation seems to rely on the concept of combinations, but there seems to be an error due to the presence of a factorial with a negative number, which is undefined. Always make sure that the combination expression is valid before attempting to calculate the probability. Remember, probability is not just about performing calculations; it’s also about making sure that the numbers used within those calculations make sense within the context of the problem being solved.

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Most popular questions from this chapter

The number of visits to a website is known to have a Poisson distribution with a mean of 8 visits per minute. a. What is the probability distribution for \(x\), the number of visits per minute? b. What is the probability that the number of visits per minute is less than or equal to \(12 ?\) c. What is the probability that the number of visits per minute is greater than \(16 ?\) d. Within what limits would you expect the number of visits to this website to lie at least \(89 \%\) of the time?

Let \(x\) be the number of successes observed in a sample of \(n=4\) items selected from a population of \(N=8 .\) Suppose that of the \(N=8\) items, \(M=5\) are considered "successes." Find the probabilities in Exercises \(8-10 .\) The probability of observing at most two successes.

In a county containing a large number of rural homes, \(60 \%\) of the homes are insured against fire. Four rural homeowners are chosen at random from this county, and \(x\) are found to be insured against fire. Find the probability distribution for \(x\). What is the probability that at least three of the four will be insured?

Let \(x\) be the number of successes observed in a sample of \(n=5\) items selected from a population of \(N=10 .\) Suppose that of the \(N=10\) items, \(M=6\) are considered "successes." Find the probabilities in Exercises \(11-13 .\) The probability of observing exactly two successes.

Draw three cards randomly from a standard deck of 52 cards and let \(x\) be the number of kings in the draw. Evaluate the probabilities and answer the questions in Exercises \(26-28\) \(P(x=3)\)

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