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Use the following information to answer the next four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu. Find the probability that at least four of the 25 patients actually have the flu.

Short Answer

Expert verified
The probability that at least four out of 25 patients have the flu is 0.0159.

Step by step solution

01

Identify the problem type

This is a binomial probability problem where we need to find the probability of having at least four patients with the flu out of 25, given the probability of a single patient having the flu is 4%, or 0.04.
02

Define the binomial distribution parameters

In a binomial distribution, the number of trials is denoted as \(n\) and the probability of success on a single trial as \(p\). For this problem, the number of trials \(n = 25\) and the probability of success \(p = 0.04\). Each patient is a trial and having the flu is considered a success.
03

Understand the probability of at least four

We are interested in finding \(P(X \geq 4)\) where \(X\) is the random variable representing the number of patients who actually have the flu. To find this probability, we can use the complement rule: \(P(X \geq 4) = 1 - P(X < 4)\).
04

Calculate the complementary probability \(P(X < 4)\)

The complementary probability \(P(X < 4)\) means finding the probability of having fewer than 4 patients with the flu, which is \(P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\). Use the binomial probability formula:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] for each value of \(k = 0, 1, 2, 3\).
05

Calculating individual probabilities

Calculate each term:1. \(P(X = 0) = \binom{25}{0} (0.04)^0 (0.96)^{25} = 0.3603 \)2. \(P(X = 1) = \binom{25}{1} (0.04)^1 (0.96)^{24} = 0.3755\)3. \(P(X = 2) = \binom{25}{2} (0.04)^2 (0.96)^{23} = 0.1888\)4. \(P(X = 3) = \binom{25}{3} (0.04)^3 (0.96)^{22} = 0.0595 \)
06

Sum the individual probabilities

Now sum these probabilities to get \(P(X < 4)\):\[ P(X < 4) = 0.3603 + 0.3755 + 0.1888 + 0.0595 = 0.9841 \]
07

Calculate \(P(X \geq 4)\)

Now find \(P(X \geq 4)\) using the complement rule:\[ P(X \geq 4) = 1 - P(X < 4) = 1 - 0.9841 = 0.0159 \]

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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 fundamental concept in probability and statistics. It models scenarios where there are two possible outcomes for each trial, commonly referred to as "success" and "failure." In our exercise, each patient calling the medical advice line either has the flu (success) or does not have the flu (failure). This distribution is characterized by two parameters: the number of trials ( ") and the probability of success on a single trial ( p").
  • The **number of trials**, denoted by , is the number of independent experiments we conduct. In the example, it is the 25 patients calling in.
  • The **probability of success** p", represents the likelihood of a single trial resulting in success. Here, it is the 4% chance that a patient truly has the flu.
The binomial distribution gives us a way to find the probability of a certain number of successes across these trials. This is particularly useful in medical statistics, quality control, and any field that involves binomial processes.
Complement Rule
The complement rule is a helpful trick in probability that allows us to find the probability of an event by considering its opposite. It states that the probability of an event happening is equal to one minus the probability of it not happening: \( P(A) = 1 - P(A^c) \). In simpler terms, if you know how likely it is for an event not to happen, you can easily find out how likely it is to happen.

In our problem, we use the complement rule to find out the probability of having at least four patients with the flu. Directly calculating **\( P(X \geq 4) \)** can be complex, but it is easier to calculate **\( P(X < 4)\)** because it is the probability of fewer than four patients having the flu, and then use the complement rule. Therefore,\( P(X \geq 4) = 1 - P(X < 4) \).
  • This approach simplifies the computation and is often easier to apply when the event of interest involves words like "at least."
  • By first calculating the simpler, complementary scenario, you can then find the more complex probability.
Probability of Success
The probability of success in a binomial experiment tells us how likely a single trial is to result in what we define as a success. This is denoted as p"). For our exercise, success is defined as a patient actually having the flu after calling the advice line.
  • **Probability of Success (\( p \))** is a fixed value for each trial. In the current scenario, it is 0.04, meaning each patient correctly having the flu is an unlikely but possible event.
  • The term "success" is relative to the context and does not necessarily imply a positive outcome. It merely refers to the occurrence of the specific event under study.
Understanding the probability of success helps in setting our expectations about the outcomes we observe, whether we're dealing with clinical trials, quality testing, or survey results.
Random Variable
A random variable is a central subject in probability theory. It's a numerical outcome of a random process or experiment. In this context, the random variable X") represents the number of patients out of the 25 who actually have the flu.
  • Random variables can be **discrete** (specific, countable outcomes) or **continuous** (any value within a range). For binomial problems, we're dealing with discrete random variables.
  • The value of the random variable reflects the possible outcomes of the experiment. Here, X=0 through X=25], which signify how many of the patients actually have the flu.
In this flu probability problem, analyzing the random variable helps quantify the uncertainty and randomness inherent in real-world data. By understanding this variable, we can calculate probabilities and make informed predictions on larger scales.

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