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The probability that a lab specimen contains high levels of contamination is \(0.10 .\) Five samples are checked, and the samples are independent. (a) What is the probability that none contain high levels of contamination? (b) What is the probability that exactly one contains high levels of contamination? (c) What is the probability that at least one contains high levels of contamination?

Short Answer

Expert verified
(a) 0.59049 (b) 0.32805 (c) 0.40951

Step by step solution

01

Identify the Probability Type

This problem is a classic example of a Binomial probability distribution, as we are dealing with independent trials (the samples) and a fixed probability for success (finding contamination).
02

Probability of No Contamination

To find the probability that none of the samples (\( n = 5 \) samples) contain high contamination, use the formula \( (1 - p)^n \), where \( p = 0.10 \). Thus, the probability is \( (1 - 0.10)^5 = 0.9^5 \).
03

Calculate \(0.9^5\)

Calculate \( 0.9^5 \) using a calculator: \( 0.9^5 = 0.59049 \). Therefore, \( P(0) = 0.59049 \).
04

Probability of Exactly One Contamination

Use the Binomial probability formula: \( P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( n = 5 \), \( k = 1 \), and \( p = 0.10 \). Thus, \( P(1) = \binom{5}{1} (0.10)^1 (0.9)^4 \).
05

Calculate \(\binom{5}{1}\), \((0.9)^4\), and Final Probability

Calculate \(\binom{5}{1} = 5\), \(0.9^4 = 0.6561\). The probability is: \( P(1) = 5 \times 0.10 \times 0.6561 = 0.32805 \).
06

Probability of At Least One Contamination

The probability of at least one containing high contamination is the complement of none being contaminated. Hence, \( P(X \geq 1) = 1 - P(0) = 1 - 0.59049 \).
07

Calculating the Complement

Subtract \( 0.59049 \) from 1 to find \( P(X \geq 1) \). Thus, \( P(X \geq 1) = 0.40951 \).

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Key Concepts

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

Independent Trials
When dealing with probability, it's important to understand the concept of independent trials. An independent trial is a scenario where the outcome of one event doesn't influence the outcome of another. Imagine flipping a coin multiple times. No matter how many times you flip it, the chance of landing on heads or tails doesn't change from one flip to the next.
This characteristic makes things like laboratory tests, such as our contamination example, predictable using statistical methods like the Binomial Probability Distribution. Each sample tested for contamination does not affect others, so they're considered independent. Because the contamination probability remains constant at 0.10 for each sample, we can apply a simple formula to determine probabilities for different contamination outcomes across multiple samples.
Probability of Contamination
The probability of contamination, in this context, is the likelihood that any given sample will show high levels of contamination. In our example, this probability is set at 0.10, or 10%. This is a fixed probability, meaning it doesn't change from one sample to the next. Therefore, for five independent samples, each tested separately, we can calculate various probabilities:
  • The probability that none of the samples are contaminated was calculated as \((1 - 0.10)^5 = 0.59049\), thanks to the independence of each sample.
  • The probability of finding exactly one contaminated sample used the Binomial formula: \(P(X = 1) = \binom{5}{1} \times (0.10)^1 \times (0.9)^4 = 0.32805\).
This concept underlines how fixed probabilities enable us to predict outcomes across several trials.
Complement Rule
The Complement Rule is a fundamental principle in probability that helps simplify calculations, especially when dealing with the probability of complex events. Essentially, it states that the probability of an event happening is 1 minus the probability of it not happening. In simpler terms, if you know the probability of something not occurring, then you can easily find out the probability of it occurring by subtracting the known probability from 1.
In our contamination problem, calculating the probability that at least one sample is contaminated was made easier using the Complement Rule. We first found the probability that no samples were contaminated, which was 0.59049. Using the complement, we calculated\(P(X \geq 1) = 1 - 0.59049 = 0.40951\). This made it much simpler than counting all possible outcomes where at least one sample might be contaminated.

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