A certain virus infects one in every 600 people. A test used to detect the virus in a person is positive 90% of the time if the person has the virus and 10% of the time if the person does not have the virus. Let A be the event "the person is infected" and B be the event "the person tests positive."
(a) Find the probability that a person has the virus given that they have tested positive.
(b) Find the probability that a person does not have the virus given that they have tested negative.
To understand the solution given below, you will need to know a few definitions or identities:
1. P(A)=probability of event A will happen.
2. definition of conditional probability:
P(A|B)=P(A∧B)/P(B)
i.e. Probability of event A happening GIVEN that B has already happened.
3. probability of a complement
P(~A)=1-P(A)
4. identity:
P(A∨B)=P(A)+P(B)-P(A∧B)
5. De Morgan's law:
~A ∧ ~B \equiv; ~(A ∨ B)
so
P(~A ∧ ~B)=1-P(A ∨ B)
We are given
P(A)=1/600;
=> P(~A)=1-1/600=599/600;
P(B|A)=0.9;
=> P(B∧A)/P(A)=0.9
=> P(B∧A)=0.9/600=11500=3/2000
P(B|~A)=0.1;
=> P(B∧~A)/P(~A)=0.1
=> P(B∧~A)=0.1*P(~A)=599/6000
This also means that
P(B∧A)+P(B∧~A)=3/2000+599/6000
=P(B∧(A∨~A)=608/6000
=P(B)
=608/6000
=38/375
=> P(~B)=1-P(B)=1-38/375=337/375
(a) find P(A|B)
P(A|B)
=P(A∧B)/P(B)
=P(B∧A)/P(B)
Substitute P(B∧A) and P(B)
to solve for P(A|B)
Hint: answer to part (a) is much greater than P(A).
(b) find P(~A|~B)
P(~A|~B)
=P(~A∧~B)/P(~B)
=(1-P(A∨B))/P(~B)
=(1-(P(A)+P(B)-P(A∧B))/P(~B)
All the quantities P(A), P(B), P(A∧B) and P(~B) have been calculated previously, so just substitute to find the answer.
Hint: answer for part B is very close to 1.
Finally, if you need detailed explanations or discussions, please post.
To solve these problems, we will use conditional probability, which is denoted as P(A|B), where A and B are events. For example, P(A|B) represents the probability of event A occurring given that event B has occurred.
(a) We want to find the probability that a person has the virus given that they have tested positive: P(A|B). We can use Bayes' theorem to calculate this probability.
Bayes' theorem states: P(A|B) = (P(B|A) * P(A)) / P(B)
In this case:
- P(A) is the probability that a person has the virus, which is given as 1/600 or approximately 0.00167.
- P(B|A) is the probability that a person tests positive given that they have the virus, which is given as 0.9 or 90%.
- P(B) is the probability that a person tests positive, regardless of whether they have the virus or not.
To find P(B), we need to consider two possibilities: a person tests positive and has the virus, and a person tests positive but does not have the virus.
Let's calculate both possibilities separately:
1. Probability of a person testing positive and having the virus:
P(B and A) = P(B|A) * P(A)
= 0.9 * 0.00167
2. Probability of a person testing positive but not having the virus:
P(B and not A) = P(B|not A) * P(not A)
The probability of a person testing positive but not having the virus is equal to the probability of testing positive (0.1 or 10%) minus the probability of testing positive and having the virus we calculated above. So,
P(B and not A) = 0.1 - (0.9 * 0.00167)
Now, we can find P(B) by summing up the probabilities of both possibilities:
P(B) = P(B and A) + P(B and not A)
Now that we have all the components, we can substitute them into Bayes' theorem formula to find P(A|B):
P(A|B) = (P(B|A) * P(A)) / P(B)
Plug in the values and calculate.
(b) Similarly, we want to find the probability that a person does not have the virus given that they have tested negative: P(not A|not B).
To solve this problem, we can use the complement rule. The complement of event A (not A) is the event that a person does not have the virus.
P(not A|not B) = 1 - P(A|not B)
Using the same method as in part (a), we can calculate P(A|not B) and then substitute the value into the formula to find P(not A|not B).
Note: If you provide specific values for P(not A) and P(not B), I can help you calculate the probabilities more precisely.