Answer :
Answer:
Part (a) The probability that the test comes back negative for all four people is: 0.98401 (approximately)
Part (b) The probability that the test comes back positive for at least one of the four people is: 0.01599 (approximately)
Step-by-step explanation:
Consider the provided information.
The test will accurately come back negative if the antibody is not present (in the test subject) 99.6% of the time.
99.6% can be written as 0.996 which is the probability of test is negative.
The probability of a test coming back positive when the antibody is not present (a false positive) is 0.004.
Part (a)
The probability that the test comes back negative for all four people is:
P(all 4 tests are negative) = (0.996)(0.996)(0.996)(0.996)
P(all 4 tests are negative) = 0.98401 (approximately)
Part (b)
The probability that the test comes back positive for at least one of the four people:
At least one means it may be 1 or 2 or 3 or all four gets a positive test. In other word it can be say that all 4 tests are negative.
As we know the probability of all tests are negative is 0.98401. So subtract this number from 1.
P(at least one of 4 people) = 1-0.98401
P(at least one of 4 people) = 0.01599 (approximately)
Using the binomial probability distribution, it is found that:
- a) 0.9841 = 98.41% probability that the test comes back negative for all four people.
- b) 0.0159 = 1.59% probability that the test comes back positive for at least one of the four people.
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For each test, there are only two possible outcomes. Either it comes back positive, or it comes back negative. The probability of a test coming back positive is independent of any other test, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of a success on a single trial.
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- Four people, thus [tex]n = 4[/tex]
- They do not have the disease, thus 0.996 probability of a negative test, which means that [tex]p = 0.996[/tex].
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Item a:
This probability is P(X = 4), thus:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 4) = C_{4,4}.(0.996)^{4}.(0.004)^{0} = 0.9841[/tex]
0.9841 = 98.41% probability that the test comes back negative for all four people.
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Item b:
This probability is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
Considering the previous item:
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.9841 = 0.0159[/tex]
0.0159 = 1.59% probability that the test comes back positive for at least one of the four people.
A similar problem is given at https://brainly.com/question/15557838