In a certain population, 1% of all individuals are carriers of a particular disease. A diagnostic test for this disease has a 90% detection rate for carriers and a 5% detection rate for non- carriers. Suppose that the diagnostic test is applied independently to two different samples from the same randomly selected individual.

a) What is the probability that both test yield the same result?
B) If both tests are positive, what is the probability that the selected individual is a carrier?

i have no idea on where to start.

Excel spreadsheets are very helpful for these kinds of problems.

Start by building a probability tree, calculate all possible values. There are 2 kinds of people (with and without), and each test has two answers (yes, no). So, there should be 8 possible outcomes.
The probability of seeing a person with the disease and both test positive is: .01*.9*.9 = .0081
of seeing a person with the disease and the first test is positive and the second negative = .01*.9*.1 = .0009
Repeat for the other 6 possibilities. If you did the calculations right, the sum of the 8 values should be 1.0.

Then you will have all the information you need.

Well, fear not, my confused friend! I'm here to help you out with some clownish reasoning on this statistical problem.

a) To determine the probability that both tests yield the same result, we can break it down into two cases: both positive or both negative. Let's calculate it step by step!

Case 1: Both tests are positive. The probability of a positive test for a carrier is 90%, and for a non-carrier is 5%. So, the probability of both tests being positive is 0.01 (carrier) * 0.01 (carrier) = 0.0001.

Case 2: Both tests are negative. The probability of a negative test for a carrier is 10% (since the test has a 90% detection rate, the complement is a 10% chance of a false negative), and for a non-carrier is 95%. So, the probability of both tests being negative is 0.99 (non-carrier) * 0.99 (non-carrier) = 0.9801.

Since we're interested in the probability of both tests yielding the same result, we add the probabilities of both cases: 0.0001 + 0.9801 = 0.9802. Thus, the probability that both tests yield the same result is approximately 0.9802, or 98.02%.

b) Now, let's tackle the second part! We want to find the probability that the individual is a carrier given that both tests are positive.

To calculate this, we'll use Bayes' theorem. The probability that the individual is a carrier, given that both tests are positive, can be found using the following formula:

Probability (Carrier | Positive) = (Probability (Positive | Carrier) * Probability (Carrier)) / Probability (Positive)

The Probability (Positive | Carrier) is the detection rate for carriers, which is 90%. Probability (Carrier) is the carrier rate in the population, which is 1%. Probability (Positive) is the probability of both tests being positive, which we calculated earlier as 0.0001.

Let's put it all together:

Probability (Carrier | Positive) = (0.90 * 0.01) / 0.0001 = 0.09 / 0.0001 = 0.09 * 10,000 = 900.

Therefore, if both tests are positive, the probability that the selected individual is a carrier is approximately 900, or 900%.

Hope that brings a smile to your statistical face!

To solve this problem, we can use conditional probability and the multiplication rule.

a) To find the probability that both tests yield the same result, we need to consider two cases: both tests are positive or both tests are negative.

Let's denote "C" as the event that the individual is a carrier and "N" as the event that the individual is a non-carrier.

Case 1: Both tests are positive.
In this case, we need to calculate the probability that both tests detect the individual as a carrier. Since the diagnostic test has a 90% detection rate for carriers, the probability of both tests detecting the individual as a carrier is 0.9 * 0.9 = 0.81.

Case 2: Both tests are negative.
In this case, we need to calculate the probability that both tests detect the individual as a non-carrier. Since the diagnostic test has a 5% detection rate for non-carriers, the probability of both tests detecting the individual as a non-carrier is 0.05 * 0.05 = 0.0025.

The probability that both tests yield the same result is the sum of the probabilities from Case 1 and Case 2:
Probability (both tests yield the same result) = Probability (Case 1) + Probability (Case 2) = 0.81 + 0.0025 = 0.8125.

b) To find the probability that the selected individual is a carrier given that both tests are positive, we need to apply Bayes' theorem.

Let's denote "P(C)" as the probability of being a carrier and "P(PP)" as the probability of both tests being positive.

According to Bayes' theorem:

P(C|PP) = (P(PP|C) * P(C)) / P(PP)

We know that:
P(PP|C) = 0.9 (probability that both tests are positive given that the individual is a carrier).
P(C) = 0.01 (probability of being a carrier).

To find P(PP), we need to calculate the probability that both tests are positive in either the carrier or non-carrier group:

P(PP) = [P(PP|C) * P(C)] + [P(PP|N) * P(N)]

P(PP|N) = 0.05 (probability that both tests are positive given that the individual is a non-carrier).

To find P(N), we subtract the probability of being a carrier from 1:
P(N) = 1 - P(C) = 1 - 0.01 = 0.99.

Now we can calculate P(PP):

P(PP) = [0.9 * 0.01] + [0.05 * 0.99] = 0.009 + 0.0495 = 0.0585.

Finally, we can calculate P(C|PP) using Bayes' theorem:

P(C|PP) = (0.9 * 0.01) / 0.0585 ≈ 0.0154.

Therefore, if both tests are positive, the probability that the selected individual is a carrier is approximately 0.0154, or 1.54%.

To solve this problem, you can use a combination of conditional probability and the rules of probability. Let's break it down step by step:

a) To find the probability that both tests yield the same result, we need to consider two possibilities: both tests are positive or both tests are negative.

Let's start with the scenario where both tests are positive. The probability of a carrier testing positive is 0.01 (1% of the population), and the probability of a non-carrier testing positive is 0.05 (5% of the population). Since these tests are independent, the probability of both events happening is the product of their individual probabilities: 0.01 * 0.05 = 0.0005.

Now, let's consider the scenario where both tests are negative. In this case, the carrier tests negative with a probability of 0.1 (100% - 90% detection rate), and the non-carrier tests negative with a probability of 0.95 (100% - 5% detection rate). Again, the probability of both events happening is their product: 0.1 * 0.95 = 0.095.

To get the probability that both tests yield the same result, we sum up the probabilities from both scenarios: 0.0005 + 0.095 = 0.0955 (or approximately 0.095).

b) If both tests are positive, we need to find the probability that the selected individual is a carrier. This is an example of conditional probability. We want to calculate the probability of the individual being a carrier given that both tests are positive.

Let's use Bayes' Theorem to express this conditional probability:

P(Carrier | both positive) = (P(both positive | Carrier) * P(Carrier)) / P(both positive)

P(both positive | Carrier) is the probability that both tests are positive given that the individual is a carrier. Since the tests are independent, this is simply the product of their individual probabilities: 0.9 * 0.9 = 0.81.

P(Carrier) is the probability of an individual being a carrier, which is given as 0.01 (1% of the population).

P(both positive) is the probability that both tests are positive, which we calculated in part (a) as 0.095.

Substituting these values into Bayes' Theorem:

P(Carrier | both positive) = (0.81 * 0.01) / 0.095 ≈ 0.0853,

So, if both tests are positive, the probability that the selected individual is a carrier is approximately 0.0853 (or 8.53%).