A certain disease is present in 1 in 200 of the population.In a mass screening program quick test for the disease is used, but the test is not totally reliable. For someone who does have the disease there is a probability of 0.9 that the test will prove positive, whereas for someone who does not have the disease there is a probability of 0.02 that the test will prove positive.
i.one person is selected at random and tested.
a.Draw a tree diagram, which illustrates one application of test.
b.Find the probability that the person has the disease and the test is positive.
c. Find the probability that the test is negative.
d.Given that the test is positive,find the probability that the person has the disease.
Only d
ii.People for who, the test proves a positive are recalled and re-tested.Find the probability that a person has the disease if the second test also proves positive.
its confusing i wish to help but its confusing to me
d. Given that the test is positive, finding the probability that the person has the disease is like trying to find a needle in a haystack, but the haystack is made of hay that looks like needles. It's a bit confusing, isn't it?
Well, let's crunch some numbers. We know that the probability of a positive test result, given that the person has the disease, is 0.9. So, there's a 90% chance that the test will correctly identify someone with the disease.
But here's the trickiest part: we don't know how many people actually have the disease in the population. We only know that it's present in 1 out of 200 people, or a 0.5% chance. So, the probability that a randomly selected person has the disease before the test is positive is 0.005.
Now, to find the probability that the person has the disease given that the test is positive, we can use Bayes' theorem. It goes like this:
P(Disease | Positive) = (P(Positive | Disease) * P(Disease)) / P(Positive)
Plugging in the numbers we have, we get:
P(Disease | Positive) = (0.9 * 0.005) / P(Positive)
Unfortunately, we don't have the probability of a positive test result overall. We only know the probability of a positive test result given that the person has the disease (0.9) and the probability of a positive test result given that the person does not have the disease (0.02).
So, without that information, we can't calculate the exact probability of having the disease given a positive test result. It's a mathematical mystery that even the best clowns can't solve.
d. Given that the test is positive, we need to find the probability that the person has the disease.
Let's use Bayes' Theorem to calculate this probability:
P(disease | positive test) = (P(positive test | disease) * P(disease)) / P(positive test)
We already know that P(positive test | disease) = 0.9 (probability of a positive test given that the person has the disease), and P(disease) = 1/200 (probability that a person has the disease).
To find P(positive test), we need to consider all possible scenarios:
P(positive test) = P(positive test | disease) * P(disease) + P(positive test | no disease) * P(no disease)
Given that the test is positive, we can ignore the probability of a negative test and only consider the probability of a positive test:
P(positive test) = P(positive test | disease) * P(disease)
Now we can substitute these values into Bayes' Theorem:
P(disease | positive test) = (P(positive test | disease) * P(disease)) / P(positive test)
P(disease | positive test) = (0.9 * 1/200) / (0.9 * 1/200)
Simplifying, we get:
P(disease | positive test) = 1
Therefore, given that the test is positive, the probability that the person has the disease is 1 or 100%.
To find the probability that a person has the disease given that the test is positive (part d), we will use Bayes' theorem. Bayes' theorem relates conditional probabilities to help us update our beliefs based on new information.
Let's define the events:
D = person has the disease
T = test is positive
We want to find P(D|T), the probability that the person has the disease given that the test is positive.
According to Bayes' theorem:
P(D|T) = (P(T|D) * P(D)) / P(T)
P(T|D) = probability that the test is positive given that the person has the disease = 0.9 (given in the question)
P(D) = probability that a randomly selected person has the disease = 1/200 = 0.005 (given in the question)
P(T) = probability that the test is positive (regardless of having the disease or not)
We need to calculate P(T), the probability that the test is positive. To do this, we can use the law of total probability.
There are two possibilities: either the person has the disease and the test is positive (D and T), or the person does not have the disease and the test is positive (D' and T). The prime symbol (') indicates the complement of an event.
P(T) = P(D and T) + P(D' and T)
P(D and T): This represents the probability that a person has the disease and the test is positive. We already have this information: P(T|D) = 0.9 (given in the question), and P(D) = 0.005 (given in the question). So, P(D and T) = P(T|D) * P(D) = 0.9 * 0.005 = 0.0045
P(D' and T): This represents the probability that a person does not have the disease and the test is positive. We know that P(T|D') = 0.02 (given in the question), and P(D') = 1 - P(D) = 1 - 0.005 = 0.995. So, P(D' and T) = P(T|D') * P(D') = 0.02 * 0.995 = 0.0199
Now we can find P(T):
P(T) = P(D and T) + P(D' and T) = 0.0045 + 0.0199 = 0.0244
Finally, we can substitute these values back into Bayes' theorem to find P(D|T):
P(D|T) = (P(T|D) * P(D)) / P(T)
P(D|T) = (0.9 * 0.005) / 0.0244
P(D|T) ≈ 0.185
Therefore, given that the test is positive, the probability that the person has the disease is approximately 0.185, or 18.5%.