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A study considered risk factors for HIV infection among intravenous drug users [11] . It found that \(40 \%\) of users who had \(\leq 100\) injections per month (light users) and \(55 \%\) of users who had \(>100\) injections per month (heavy users) were HIV positive. What is the probability that exactly 3 of 5 light users are HIV positive?

Short Answer

Expert verified
The probability is 0.2304.

Step by step solution

01

Understand the Problem

We want to find the probability of 3 out of 5 light users being HIV positive. Each light user has a 40% chance of being HIV positive.
02

Identify the Mathematical Model

This problem follows a binomial distribution because there are a fixed number of trials (5 light users), each with two outcomes (HIV positive or not), and the probability of being HIV positive (0.4) is the same for each user.
03

Apply the Binomial Probability Formula

The probability of exactly 3 successes (HIV positive users) in 5 trials is given by the binomial probability formula: \[ P(X = k) = \binom{n}{k} (p^k) ((1-p)^{n-k}) \]Where \(n = 5\), \(k = 3\), and \(p = 0.4\).
04

Calculate the Binomial Coefficient

Calculate \(\binom{5}{3}\) as follows:\[\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10\]
05

Compute the Probability

Substitute the values into the formula:\[ P(X = 3) = 10 \times (0.4)^3 \times (0.6)^2 \]Calculate it step by step:
06

Simplify the Expression

First calculate \((0.4)^3 = 0.064\) and \((0.6)^2 = 0.36\).
07

Final Calculation

Now, multiply all parts together: \[ P(X = 3) = 10 \times 0.064 \times 0.36 = 0.2304\]

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Key Concepts

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

HIV Risk Factors
Understanding HIV risk factors is crucial when studying how this virus spreads among different populations. HIV, or Human Immunodeficiency Virus, is primarily transmitted through certain body fluids like blood, semen, and vaginal fluids. Intravenous drug users represent a group particularly at risk due to needle sharing practices.

In the scenario provided, individuals who engage in drug use through injections are categorized based on their frequency of usage. Light users are classified as individuals who inject drugs fewer than or equal to 100 times a month, while heavy users inject more frequently. Each of these practices carries a different statistical risk for contracting HIV.

The study cited indicates that 40% of light users and 55% of heavy users are HIV positive. This variation highlights how increased frequency in risky behavior correlates with a higher chance of infection, important for developing targeted interventions.
Binomial Distribution
The binomial distribution is a common probability distribution that models the number of successful outcomes in a fixed number of trials. Each trial has two possible outcomes, such as success or failure. In our context, a success is defined as a light drug user testing positive for HIV.

To use binomial distribution, certain criteria must be met:
  • There must be a fixed number of trials, which here is 5 users.
  • Each trial is independent, meaning one user's status doesn’t affect another's.
  • The probability of success (being HIV positive) remains constant, here 0.4 (or 40%) for all light users.

This setup allows us to use the binomial probability formula to calculate exact probabilities of a given number of successes, such as 3 out of 5 users being HIV positive.
Probability Calculation
Probability calculations help us quantify the chance of an event occurring based on known data. When using the binomial distribution formula, several steps are involved in finding this likelihood.

The Binomial Probability Formula is expressed as:\[ P(X = k) = \binom{n}{k} (p^k) ((1-p)^{n-k}) \]Here,
  • \(n\) is the total number of trials (5 in our case).
  • \(k\) represents the desired number of successful outcomes (3 users testing positive).
  • \(p\) is the probability of success in an individual trial (0.4 for each user).

Steps to calculate the probability:
  • First, compute the binomial coefficient \(\binom{5}{3}\), which is the number of ways to choose \(k\) successes out of \(n\) trials.
  • Next, raise the probability of success \(p\) to the power of \(k\), multiplying it with the probability of failure \((1-p)\) raised to the power \((n-k)\).
  • Finally, multiply these values together to get the probability \(P(X = 3)\), which results in approximately 0.2304 or 23.04% chance.
The use of the binomial probability formula provides a structured way to understand and compute the likelihood of events in similar scenarios.

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