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.
Your first stage had two branches, I labeled mine D and ND (disease and not disease)
Each of those has two branches which I labeled P and N (positive and negative)
The D branch is .005 and the ND is .995
The D-P branch is .9 and the D-N is .1
the ND-P is .02 and the ND-N is .98
So now you can find the prob of any combination, e.g.
"Find the probability that the person has the disease and the test is positive"
-----> that would be the path D,D-P
= (.005)(.9) = .0045
do no. d
a. To draw a tree diagram, we need to consider the possible outcomes at each step.
Step 1: Whether the person has the disease or not
- Disease (1/200)
- No disease (199/200)
Step 2: Test result
- Positive (for both disease and no disease)
- Negative (for both disease and no disease)
So, the tree diagram would look like this:
Disease (1/200)
/
/
Positive (0.9)
/
/
/
/
/
/
/
/
/
/
/
Person
\
\
\
\
\
No Disease (199/200)
\
\
\
\
\
Negative (0.02)
\
\
\
\
\
b. To find the probability that the person has the disease and the test is positive, we need to multiply the probabilities along the path leading to this outcome.
P(Has disease and test is positive) = P(Has disease) * P(Test is positive | Has disease)
P(Has disease and test is positive) = (1/200) * 0.9
P(Has disease and test is positive) = 0.0045
Hence, the probability that the person has the disease and the test is positive is 0.0045.
c. The probability that the test is negative can be found by subtracting the probability that the test is positive from 1.
P(Test is negative) = 1 - P(Test is positive)
P(Test is negative) = 1 - 0.0045
P(Test is negative) = 0.9955
Hence, the probability that the test is negative is 0.9955.
d. To find the probability that the person has the disease given that the test is positive, we need to use Bayes' theorem.
P(Person has disease | Test is positive) = (P(Test is positive | Person has disease) * P(Person has disease)) / P(Test is positive)
P(Person has disease | Test is positive) = (0.9 * (1/200)) / (0.0045)
P(Person has disease | Test is positive) = 0.9 / 0.0045
P(Person has disease | Test is positive) = 200
Hence, given that the test is positive, the probability that the person has the disease is 200.
a. To draw a tree diagram for this scenario, we can follow these steps:
Step 1: Start by drawing a root node at the top of the diagram, representing the initial selection of a person.
Step 2: From the root node, draw two branches representing the two possible outcomes: having the disease (D) or not having the disease (D').
Step 3: From each branch, draw two more branches representing the possible outcomes of the test: positive (P) or negative (P').
Step 4: Label the probabilities on each branch. For the having disease branch, the probability is 1/200 (1 in 200). For the not having disease branch, the probability is 199/200 (199 in 200). For the positive test outcome branches, the probabilities are 0.9 and 0.02, respectively. For the negative test outcome branches, the remaining probabilities can be calculated as 1 minus the corresponding positive test outcome probabilities.
The tree diagram should now display all the possible outcomes and their associated probabilities.
b. To find the probability that the person has the disease and the test is positive, we need to consider the two branches that lead to a positive test outcome: having the disease and not having the disease, multiplied by their respective probabilities. So the probability can be calculated as:
Probability(Disease and Positive Test) = Probability(Disease) * Probability(Positive Test | Disease) = (1/200) * 0.9
c. To find the probability that the test is negative, we need to consider the two branches that lead to a negative test outcome: having the disease and not having the disease, multiplied by their respective probabilities. So the probability can be calculated as:
Probability(Negative Test) = Probability(Disease) * Probability(Negative Test | Disease) + Probability(Not Having Disease) * Probability(Negative Test | Not Having Disease) = (1/200) * (1 - 0.9) + (199/200) * (1 - 0.02)
d. Given that the test is positive, we need to find the probability that the person has the disease. This can be calculated using Bayes' theorem:
Probability(Disease | Positive Test) = (Probability(Disease) * Probability(Positive Test | Disease)) / Probability(Positive Test)
We already have the values for Probability(Disease), Probability(Positive Test | Disease), and Probability(Positive Test) from the previous calculations.