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Chapter 5: Problem 5

Given a binomial experiment with probability of success on a single trial \(p=0.40,\) find the probability that the first success occurs on trial number \(n=3\)

Short Answer

Expert verified
The probability of the first success on trial 3 is 0.144.

Step by step solution

01

Identify the Distribution Type

Since we are looking for the probability of the first success occurring on trial number 3, we will use the geometric distribution. The geometric distribution models the number of trials needed for the first success in a series of independent and identically distributed Bernoulli trials.
02

Write the Geometric Probability Formula

The probability that the first success occurs on trial number is given by the formula: \(P(X = n) = (1-p)^{n-1} \cdot p\) where \(p\) is the probability of success on a single trial.
03

Substitute the Given Values into the Formula

We are given \(p = 0.40\) and \(n = 3\). Substitute these into the formula: \(P(X = 3) = (1 - 0.40)^{3-1} \times 0.40\).
04

Calculate the Probability

First, compute \(1 - 0.40 = 0.60\). Then, calculate: \(P(X = 3) = 0.60^{2} \times 0.40 = 0.36 \times 0.40 = 0.144\). Thus, the probability that the first success occurs on trial number 3 is 0.144.

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Key Concepts

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

Probability of Success
The probability of success in a geometric distribution refers to the likelihood of achieving a successful outcome in a single trial of an experiment. In this context, a "success" is defined based on the specifics of the experiment itself. It's essential to clearly understand what a success represents before solving problems related to probability.

For our exercise, the probability of success (p = 0.40) means there is a 40% chance that any given trial will result in success. As the probability of success directly influences the distribution, knowing this value is crucial. Since each trial is independent, the probability remains constant for each trial.

  • The probability of success (p) defines the likelihood of a successful trial.
  • It is a central parameter in geometric distribution calculations.
  • For our problem, it's fixed at 0.40 or 40%.
Bernoulli Trials
A Bernoulli trial is a random experiment with exactly two possible outcomes: "success" and "failure." Every Bernoulli trial is independent, meaning the result of one trial does not affect another.

In a geometric distribution framework, a series of independent Bernoulli trials is conducted to determine how many trials are needed to achieve the first success. In our example, each trial, like flipping a coin or rolling a dice, will result in either a success (defined here as a probability of 0.40) or a failure.

  • Bernoulli trials have only two outcomes.
  • They are independent of each other.
  • Success probability remains constant for each trial.
Probability Formula
The geometric distribution's probability formula helps us determine the likelihood of the first success occurring on a specific trial. For a geometric distribution, the probability that the first success occurs on the nth trial is given by: \[ P(X = n) = (1-p)^{n-1} \cdot p \] Where - \( p \) is the probability of success on any given trial, and - \( n \) is the trial number at which the first success occurs.

In the problem provided, substituting the given values (\( p = 0.40 \) and \( n = 3 \)), we apply the formula to calculate the probability:

\[ P(X = 3) = (1 - 0.40)^{3-1} \times 0.40 = 0.60^{2} \times 0.40 = 0.144 \]

This result indicates that there is a 14.4% chance that the first success will occur on the third trial.

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Most popular questions from this chapter

Suppose we have a binomial experiment, and the probability of success on a single trial is 0.02. If there are 150 trials, is it appropriate to use the Poisson distribution to approximate the probability of three successes? Explain.

Have you ever tried to get out of jury duty? About \(25 \%\) of those called will find an excuse (work, poor health, travel out of town, etc.) to avoid jury duty (Source: Bernice Kanner, Are You Normal?, St. Martin's Press). (a) If 12 people are called for jury duty, what is the probability that all 12 will be available to serve on the jury? (b) If 12 people are called for jury duty, what is the probability that 6 or more will not be available to serve on the jury? (c) Find the expected number of those available to serve on the jury. What is the standard deviation? (d) How many people \(n\) must the jury commissioner contact to be \(95.9 \%\) sure of finding at least 12 people who are available to serve?

What is the age distribution of promotion-sensitive shoppers? A supermarket super shopper is defined as a shopper for whom at least \(70 \%\) of the items purchased were on sale or purchased with a coupon. The following table is based on information taken from Trends in the United States (Food Marketing Institute, Washington, D.C.). $$\begin{array}{|l|ccccc|} \hline \text { Age range, years } & 18-28 & 29-39 & 40-50 & 51-61 & 62 \text { and over } \\\ \hline \text { Midpoint } x & 23 & 34 & 45 & 56 & 67 \\ \hline \begin{array}{l} \text { Percent of } \\ \text { super shoppers } \end{array} & 7 \% & 44 \% & 24 \% & 14 \% & 11 \% \\ \hline \end{array}$$ For the 62 -and-over group, use the midpoint 67 years. (a) Using the age midpoints \(x\) and the percentage of super shoppers, do we have a valid probability distribution? Explain. (b) Use a histogram to graph the probability distribution of part (a). (c) Compute the expected age \(\mu\) of a super shopper. (d) Compute the standard deviation \(\sigma\) for ages of super shoppers.

USA Today reported that for all airlines, the number of lost bags was May: 6.02 per 1000 passengers December: 12.78 per 1000 passengers Note: A passenger could lose more than one bag. (a) Let \(r=\) number of bags lost per 1000 passengers in May. Explain why the Poisson distribution would be a good choice for the random variable \(r\) What is \(\lambda\) to the nearest tenth? (b) In the month of May, what is the probability that out of 1000 passengers, no bags are lost? that 3 or more bags are lost? that 6 or more bags are lost? (c) In the month of December, what is the probability that out of 1000 passengers, no bags are lost? that 6 or more bags are lost? that 12 or more bags are lost? (Round \(\lambda\) to the nearest whole number.)

Consider a binomial experiment with \(n=6\) trials where the probability of success on a single trial is \(p=0.85\) (a) Find \(P(r \leq 1)\) (b) Interpretation If you conducted the experiment and got fewer than 2 successcs, would you be surpriscd? Why?
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