There is a new diagnostic test for a disease that occurs in about 0.05% of the population. The test is not perfect but will detect a person with the disease 99% of the time. It wil... There is a new diagnostic test for a disease that occurs in about 0.05% of the population. The test is not perfect but will detect a person with the disease 99% of the time. It will, however, say that a person without the disease has the disease about 3% of the time. A person is selected at random from the population and the test indicates that this person has the disease. What are the conditional probabilities that (i) the person has the disease? (ii) the person does not have the disease? i) X~N(mean=3, variance=2), X~N(mean=1, variance=5) and are independent. Find bound on P[X3Y<9]. ii) Examine whether the Weak law of large numbers (WLLN) holds for sequence of independent r.v.s Xk having probability distribution P[Xk=±√(2k1)]=0.5. The M.G.F of (X,Y) is given by (0.2e^t1 + 0.5e^t2 + 0.3t^3). i) Write the probability distribution of i) (X,Y), ii) Y given X=2 iii) X given Y=2 ii) Find E(YX=2) and E(XY=2) iii) Obtain P(X+Y>5). A random sample of size six with observations 2.24, 1.97, 2.11, 1.09, 1.3 and 2.55 is drawn from population with p.d.f as follows: f(x,θ) = e^(x/θ), x>θ and θ>0 = 0; otherwise. Obtain an unbiased estimate of θ based on first order statistic and standard error of your estimate.
Understand the Problem
The question presents several statistical and probabilistic problems involving diagnostic tests, probability distributions, and estimation methods. It requires applying concepts from probability theory and statistics to solve different scenarios related to a diagnostic test, random variables, momentgenerating functions, and sample estimates.
Answer
$P(D  T^+) \approx 0.0162$, $P(\neg D  T^+) \approx 0.9838$.
Answer for screen readers
The conditional probabilities are:

The probability that the person has the disease given a positive test result is approximately $P(D  T^+) \approx 0.0162$.

The probability that the person does not have the disease given a positive test result is approximately $P(\neg D  T^+) \approx 0.9838$.
Steps to Solve

Understanding the Diagnostic Test Problem
The probability of having the disease, $P(D)$, is 0.0005, and the probability of obtaining a positive test result given the person has the disease, $P(T^+  D)$, is 0.99. Conversely, the probability of a false positive (saying the person has the disease when they do not) is $P(T^+  \neg D) = 0.03$.

Finding Conditional Probabilities
Using Bayes' theorem: [ P(D  T^+) = \frac{P(T^+  D) \cdot P(D)}{P(T^+)} ] First, we need to calculate $P(T^+)$: [ P(T^+) = P(T^+  D) \cdot P(D) + P(T^+  \neg D) \cdot P(\neg D) ] [ P(\neg D) = 1  P(D) = 1  0.0005 = 0.9995 ]
Substitute the values: [ P(T^+) = 0.99 \cdot 0.0005 + 0.03 \cdot 0.9995 ]

Calculating $P(T^+)$
Now perform the calculations for ( P(T^+) ): [ P(T^+) = 0.000495 + 0.029985 = 0.03048 ]

Compute $P(D  T^+)$
Now substitute values back into Bayes' theorem: [ P(D  T^+) = \frac{0.99 \cdot 0.0005}{0.03048} ]
Conduct the calculation: [ P(D  T^+) = \frac{0.000495}{0.03048} \approx 0.0162 ]

Finding $P(\neg D  T^+)$
The probability that the person does not have the disease given a positive test result is: [ P(\neg D  T^+) = \frac{P(T^+  \neg D) \cdot P(\neg D)}{P(T^+)} ] Substitute values: [ P(\neg D  T^+) = \frac{0.03 \cdot 0.9995}{0.03048} ]
Calculate: [ P(\neg D  T^+) \approx \frac{0.029985}{0.03048} \approx 0.9838 ]
The conditional probabilities are:

The probability that the person has the disease given a positive test result is approximately $P(D  T^+) \approx 0.0162$.

The probability that the person does not have the disease given a positive test result is approximately $P(\neg D  T^+) \approx 0.9838$.
More Information
This analysis highlights the issue of false positives in diagnostic testing, which can lead to a high probability of not having the disease despite a positive test result. This scenario illustrates the crucial need to understand how diagnostic tests operate in the context of the prevalence of the disease.
Tips
 Miscalculating conditional probabilities: Ensure that Bayes' theorem is applied correctly and doublecheck the values used.
 Neglecting the base probability: Always account for the prevalence of the condition in the population when calculating probabilities.