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A chemist is interested in determining whether a certain trace impurity is present in a product. An experiment has a probability of \(.80\) of detecting the impurity if it is present. The probability of not detecting the impurity if it is absent is \(.90\). The prior probabilities of the impurity being present and being absent are \(.40\) and \(.60\), respectively. Three separate experiments result in only two detections. What is the posterior probability that the impurity is present?

Short Answer

Expert verified
Approximately 0.9041.

Step by step solution

01

Understand the Problem

We need to determine the posterior probability of the impurity being present given that two out of three experiments detect the impurity. This involves using Bayes' theorem.
02

Define the Probabilities

Let's denote the detection of impurity as 'D' and the non-detection as 'ND'. Then, the known probabilities are:- Probability of detecting the impurity when present: \( P(D|I) = 0.80 \).- Probability of non-detecting the impurity when absent: \( P(ND|A) = 0.90 \).- Prior probability of impurity being present: \( P(I) = 0.40 \).- Prior probability of impurity being absent: \( P(A) = 0.60 \).
03

Calculate the Likelihoods

We need to find the likelihood of observing two detections out of three experiments. For each scenario (impurity present or absent), we calculate:- Probability of two detections if impurity is present: \( P( ext{2 Detections}|I) = \binom{3}{2} (0.80)^2 (0.20)^1 = 3 \times 0.64 \times 0.20 = 0.384 \).- Probability of two detections if impurity is absent: \( P( ext{2 Detections}|A) = \binom{3}{2} (0.10)^2 (0.90)^1 = 3 \times 0.01 \times 0.90 = 0.027 \).
04

Use Bayes' Theorem to Find the Posterior Probability

The posterior probability that the impurity is present given two detections is calculated using Bayes' theorem:\[ P(I| ext{2 Detections}) = \frac{P( ext{2 Detections}|I) P(I)}{P( ext{2 Detections})} \]Where the total probability of two detections, \( P( ext{2 Detections}) \), is:\[ P( ext{2 Detections}) = P( ext{2 Detections}|I) P(I) + P( ext{2 Detections}|A) P(A) \]Calculate these values:- \( P( ext{2 Detections}) = 0.384 \times 0.40 + 0.027 \times 0.60 = 0.1536 + 0.0162 = 0.1698 \).Now, calculate the posterior probability:\[ P(I| ext{2 Detections}) = \frac{0.384 \times 0.40}{0.1698} = \frac{0.1536}{0.1698} \approx 0.9041 \].
05

Conclusion

The posterior probability that the impurity is present given that two detections were made out of three experiments is approximately 0.9041.

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Key Concepts

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

Posterior Probability
The posterior probability is the probability of an event occurring, given new evidence or information. In Bayesian probability, we use it to revise our beliefs about the likelihood of an event. This probability is what we aim to find in the experiment conducted by the chemist. After the results of the three experiments, with two detections of the trace impurity, the posterior probability tells us how likely it is that the impurity is actually present based on this new evidence.

In the provided example, we calculate the posterior probability of the impurity being present after conducting the experiments. The value we obtain (\( \approx 0.9041 \)) suggests a high likelihood that the impurity exists in the samples tested.
Likelihood
Likelihood refers to the probability of observing the data given a particular hypothesis. In the context of Bayesian probability, it helps assess how likely the observed data is if our initial assumptions about an event (hypothesis) are true. It's an essential part of calculating the posterior probability because it conveys how well the hypothesis explains the observed data.

For instance, in the chemist's experiment:
  • The likelihood of observing two detections if the impurity is present was calculated as \( 0.384 \).
  • The likelihood of observing the same results if the impurity is absent was much lower at \( 0.027 \).
These likelihoods indicate that the observed outcome (two detections out of three attempts) is much more consistent with the impurity being present than absent.
Bayes' Theorem
Bayes' theorem is a mathematical formula used to update probabilities based on new data or evidence. It links the prior probability and likelihood to find the posterior probability. It is crucial for revising predictive models based on new evidence.

This theorem is structured as follows:\[P(H|E) = \frac{P(E|H) \times P(H)}{P(E)}\]
  • \( P(H|E) \) represents the posterior probability.
  • \( P(E|H) \) is the likelihood of evidence \( E \) given hypothesis \( H \).
  • \( P(H) \) is the prior probability of the hypothesis.
  • \( P(E) \) is the marginal likelihood, or the total probability of the evidence.
In our chemical test example, Bayes' theorem was applied to deduce the probability of the impurity being present after the experiments, ultimately arriving at a posterior probability of approximately \( 0.9041 \).
Prior Probability
Prior probability represents our initial belief about the likelihood of a particular event before considering any new evidence. It reflects what is known about the situation before observing any data.

In the chemist's exercise, the prior probabilities were:
  • \( P(I) = 0.40 \) for the impurity being present.
  • \( P(A) = 0.60 \) for the impurity being absent.
These values set the stage for further analysis using Bayes' theorem. They provided the baseline assumptions necessary to incorporate new data (experimental results) to update the belief about the impurity's presence. The prior probabilities help in calculating how the new evidence affects our understanding of whether the impurity is present.

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