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Chapter 5: Problem 14

A research team at Cornell University conducted a study showing that approximately \(10 \%\) of all businessmen who wear ties wear them so tightly that they actually reduce blood flow to the brain, diminishing cerebral functions (Source: Chances: Risk and Odds in Everyday Life, by James Burke). At a board meeting of 20 businessmen, all of whom wear ties, what is the probability that (a) at least one tie is too tight? (b) more than two ties are too tight? (c) no tie is too tight? (d) at least 18 ties are not too tight?

Short Answer

Expert verified
(a) Use \(1-(0.90)^{20}\), (b) Use \(1-P(X \leq 2)\), (c) Use \((0.90)^{20}\), (d) Use \(P(X \leq 2)\).

Step by step solution

01

Identify Key Variables

We are given a probability problem involving businessmen wearing tight ties. Let the probability that a tie is too tight be \( p = 0.10 \). There are 20 businessmen, so \( n = 20 \). This situation can be modeled using a binomial distribution \( X \sim \text{Binomial}(n, p) \).
02

Calculating Probability for No Tight Ties

To find the probability that no tie is too tight, we use the binomial probability formula: \( P(X = 0) = \binom{20}{0} (0.10)^0 (0.90)^{20} = (0.90)^{20} \). Compute \((0.90)^{20}\).
03

Finding Probability for At Least One Tight Tie

Probability of at least one tie being too tight is the complement of no tie being too tight. Use the formula: \( P(X \geq 1) = 1 - P(X = 0) \). Insert the result from Step 2 to get \( P(X \geq 1) = 1 - (0.90)^{20} \).
04

Computing Probability for More Than Two Tight Ties

We need to calculate \( P(X > 2) = 1 - P(X \leq 2) \). Sum up the probabilities \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \) using the binomial formula and subtract from 1.
05

Use Cumulative Probabilities

Calculate \( P(X \leq 2) \) using \( P(X = 0) = (0.90)^{20} \), \( P(X = 1) = \binom{20}{1} (0.10)^1 (0.90)^{19} \) and \( P(X = 2) = \binom{20}{2} (0.10)^2 (0.90)^{18} \). Sum them up to find \( P(X \leq 2) \).
06

Calculate At Least 18 Ties Not Too Tight

Probability that at least 18 ties are not too tight is equivalent to \( P(X \leq 2) \), which we already calculated for Step 5. This is because if 18 are not too tight, at most 2 can be too tight.

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Key Concepts

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

Binomial Distribution
The binomial distribution is a common way to describe the number of successful outcomes in a fixed number of independent trials, where the outcome of each trial can either be a success or a failure. In this context, a **success** is defined as a tie being too tight; however, it's important to note that this use of "success" merely denotes the event of interest. Key features of a binomial distribution include:
  • The number of trials, denoted as \( n \). In the given problem, there are 20 trials, corresponding to the 20 businessmen.
  • The probability of success on each trial, denoted as \( p \). For a tie being too tight, \( p = 0.10 \).
  • The number of successes in the \( n \) trials, which is a random variable \( X \), following the distribution \( X \sim \text{Binomial}(n, p) \).
These elements help in determining probabilities for specific scenarios, such as at least one tie being too tight.
Complement Rule
The complement rule is a crucial concept in probability, often employed to simplify calculations in scenarios where calculating directly is cumbersome. It states that the probability of an event happening is 1 minus the probability of the event not happening.In mathematical terms:
  • If \( P(A) \) is the probability that event \( A \) happens, then \( P(A^c) = 1 - P(A) \) is the probability that event \( A \) doesn't happen.
For the stated problem, the probability of at least one tie being too tight can be calculated using its complement: the probability that no tie is too tight. This is shown by:\[P(X \geq 1) = 1 - P(X = 0)\]By computing \((0.90)^{20}\) for \(P(X = 0)\), you can easily determine \(P(X \geq 1)\). This method, using the complement rule, significantly reduces the complexity of the calculations.
Cumulative Probability
Cumulative probability refers to the probability that a random variable is less than or equal to a certain value. This is particularly useful when you want to find the likelihood of the occurrence of certain ranges of outcomes.For instance, if you want to determine the probability that two or fewer ties are too tight (i.e., \(X \leq 2\)), you calculate:\[P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\]Calculating each probability component:
  • \( P(X = 1) = \binom{20}{1} (0.10)^1 (0.90)^{19} \)
  • \( P(X = 2) = \binom{20}{2} (0.10)^2 (0.90)^{18} \)
Summing these probabilities gives the cumulative probability for ties being too tight in a range of 0 to 2. Utilizing cumulative probabilities simplifies answering "at most" and "at least" types of questions.
Binomial Probability Formula
The binomial probability formula is fundamental to determine the likelihood of observing exactly \( k \) successes in \( n \) independent trials, each with a success probability \( p \). The formula is expressed as:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]Here:
  • \( \binom{n}{k} \) represents the binomial coefficient, or the number of ways to choose \( k \) successes out of \( n \) trials.
  • \( p^k \) denotes the probability of having \( k \) successes.
  • \( (1-p)^{n-k} \) accounts for the probability of the remaining \( n-k \) trials being failures.
This formula allows the calculation of probabilities for any specific count of successes, such as no ties too tight, more than two tight ties, or any other specific scenario based on the problem setup. The binomial formula effectively utilizes the defined parameters \( n \) and \( p \) to solve specific probability queries efficiently.

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Most popular questions from this chapter

Sociologists say that \(90 \%\) of married women claim that their husband's mother is the biggest bone of contention in their marriages (sex and money are lower-rated areas of contention). (See the source in Problem 12.) Suppose that six married women are having coffee together one morning. What is the probability that (a) all of them dislike their mother-in-law? (b) none of them dislike their mother-in-law? (c) at least four of them dislike their mother-in-law? (d) no more than three of them dislike their mother-in-law?

Harper's Index reported that the number of (Orange County, California) convicted drunk drivers whose sentence included a tour of the morgue was 569 , of which only 1 became a repeat offender. (a) Suppose that of 1000 newly convicted drunk drivers, all were required to take a tour of the morgue. Let us assume that the probability of a repeat offender is still \(p=1 / 569 .\) Explain why the Poisson approximation to the binomial would be a good choice for \(r=\) number of repeat offenders out of 1000 convicted drunk drivers who toured the morgue. What is \(\lambda\) to the nearest tenth? (b) What is the probability that \(r=0\) ? (c) What is the probability that \(r>1\) ? (d) What is the probability that \(r>2\) ? (e) What is the probability that \(r>3\) ?

Which of the following are continuous variables, and which are discrete? (a) Number of traffic fatalities per year in the state of Florida (b) Distance a golf ball travels after being hit with a driver (c) Time required to drive from home to college on any given day (d) Number of ships in Pearl Harbor on any given day (e) Your weight before breakfast each morning

An archaeological excavation at Burnt Mesa Pueblo showed that about \(10 \%\) of the flaked stone objects were finished arrow points (Source: Bandelier Archaeological Excavation Project: Summer 1990 Excavations at Burnt Mesa Pueblo, edited by Kohler, Washington State University). How many flaked stone objects need to be found to be \(90 \%\) sure that at least one is a finished arrow point? (Hint: Use a calculator and note that \(P(r \geq 1) \geq 0.90\) is equivalent to \(1-P(0) \geq 0.90\), or \(P(0) \leq 0.10 .)\)

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.
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