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Use the formula \(C(n, x) p^{x} q^{n-x}\) to determine the probability of the given event. The probability of no successful outcomes in six trials of a binomial experiment in which \(p=\frac{1}{3}\)

Short Answer

Expert verified
The probability of no successful outcomes in six trials of a binomial experiment in which \(p=\frac{1}{3}\) is \(\frac{64}{729}\).

Step by step solution

01

Find the probability of failure

Since there are only two outcomes in a binomial experiment (success and failure), the probabilities of success and failure must add up to 1. In this case, we have the probability of success as \(p=\frac{1}{3}\), so we can find the probability of failure \(q\) by subtracting \(p\) from 1: \(q=1-p=1-\frac{1}{3}=\frac{2}{3}\). Thus, the probability of failure is \(q=\frac{2}{3}\).
02

Calculate the binomial coefficient

We will now calculate the binomial coefficient \(C(n, x)\), which is given by the formula: \(C(n, x)=\frac{n!}{x!(n-x)!}\). In this case, \(n=6\) and \(x=0\). Plugging these values into the formula, we get: \(C(6, 0)=\frac{6!}{0!(6-0)!}=\frac{6!}{0!6!}\). Since \(0!=1\), we can simplify this to: \(C(6, 0)=\frac{6!}{6!}=1\).
03

Apply the binomial probability formula

We now have all the values needed to apply the binomial probability formula. We are given \(p=\frac{1}{3}\) and we found \(q=\frac{2}{3}\). We also calculated the binomial coefficient as \(C(6, 0)=1\). Using the binomial probability formula \(C(n, x)p^xq^{n-x}\), we can find the probability of no successful outcomes: \(P(X=0) = C(6, 0)\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^{6-0} = 1\cdot1\cdot\left(\frac{2}{3}\right)^6\). Now we simply compute the probability: \(P(X=0) = \left(\frac{2}{3}\right)^6=\frac{64}{729}\). Therefore, the probability of no successful outcomes in the given binomial experiment is \(\frac{64}{729}\).

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Key Concepts

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

Binomial Experiment
A binomial experiment is a statistical process that involves a fixed number of trials, each with only two possible outcomes: success or failure. These trials are identical, and the probability of success is constant for each trial.

It's essential for students to understand that each trial is independent, meaning the outcome of one trial won't affect another. This is what differentiates binomial experiments from other types of probability experiments.

Let's look at an example: Tossing a fair coin ten times. Each toss could end in success (for heads) or failure (for tails), and every toss is independent of the others. Here, our binomial experiment is clearly set with 10 trials, and we have a probability of success fixed at 0.5 for each toss.

So, in the context of the original exercise, we had six trials, and the focus was on calculating the probability of zero successes. Remembering these principles about the independence and fixed probabilities is crucial for identifying and understanding binomial experiments.
Probability of Failure
The probability of failure in a binomial experiment is derived from the probability of success. Let's denote the probability of success as "p". The probability of failure, which we commonly refer to as "q," is simply the complement of the probability of success.

Mathematically, this is expressed as:
  • \( q = 1 - p \).


For instance, in a bag of colored balls where the probability of drawing a red ball (success) is 0.7, the probability of failure (drawing any other color) would be 0.3, because \( q = 1 - 0.7 = 0.3 \). It's a straightforward calculation but understanding this concept is essential, as it sets the foundation for determining binomial probabilities.

In the original exercise, we calculated "q" as \( \frac{2}{3} \), since "p" was \( \frac{1}{3} \), demonstrating how easy it is to compute "q" once "p" is known.
Binomial Coefficient
The binomial coefficient is a crucial part of the binomial probability formula and is often represented as \(C(n, x)\). It calculates the number of ways to choose "x" successes in "n" trials without regard to order.

Mathematically, it's expressed as:
  • \( C(n, x) = \frac{n!}{x!(n-x)!} \).


A common way to think about it is considering how many different combinations exist for a specific number of successes in a sequence of trials. For example, if you wanted to know how many ways you can get 2 heads in 3 coin tosses, you'd use the binomial coefficient to calculate this.

Back to the example in the original solution, we determined \( C(6, 0) = 1 \), which simplifies to calculating the probability of having zero successes in six trials. This simplification often occurs when "x" is zero or equal to "n" because the mathematics resolve to needing only one way to choose either all or none.

Thus, understanding the binomial coefficient's computation is not only about knowing factorials but visualizing how different outcomes can combine in a binomial experiment.

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Most popular questions from this chapter

ENGINE FALURES The probability that an airplane engine will fail in a transcontinental flight is .001. Assuming that engine failures are independent of each other, what is the probability that, on a certain transcontinental flight, a fourengine plane will experience a. Exactly one engine failure? b. Exactly two engine failures? c. More than two engine failures? (Note: In this event, the airplane will crash.)

A probability distribution has a mean of 42 and a standard deviation of 2 . Use Chebychev's inequality to find a bound on the probability that an outcome of the experiment lies between a. 38 and 46 . b. 32 and 52 .

QuAury CoNTRoL As part of its quality-control program, the video-game DVDs produced by Starr Communications are subjected to a final inspection before shipment. A sample of six DVDs is selected at random from each lot of DVDs produced, and the lot is rejected if the sample contains one or more defective DVDs. If \(1.5 \%\) of the DVDs produced by Starr is defective, find the probability that a shipment will be accepted.

The number of married men (in thousands) between the ages of 20 and 44 in the United States in 1998 is given in the following table: $$ \begin{array}{lccccc} \hline \text { Age } & 20-24 & 25-29 & 30-34 & 35-39 & 40-44 \\ \hline \text { Men } & 1332 & 4219 & 6345 & 7598 & 7633 \\ \hline \end{array} $$ Find the mean and the standard deviation of the given data.

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