Let's break down this HIVIV testing scenario using probabilities. We can define the
following events:
●
●
●
●
HIV: The event that a patient is infected with HIVIV.
Not-HIV The event that a patient is not infected with HIVIV.
P: The event that a patient tests positive for HIVIV.
Ng: The event that a patient tests negative for HIVIV.
From the problem statement, we have the following probabilities:
● P(HIV)=0.08 (8% of patients are infected with HIVIV)
● P(P∣HIV)=0.98 (98% of HIVIV-infected patients test positive)
● P(P∣N)=0.03 (3% of HIVIV-uninfected patients test positive)
From these, we can also deduce:
● P(N)=1−P(HIV)=1−0.08=0.92 (92% of patients are not infected with HIVIV)
● P(Ng∣HIV)=1−P(P∣HIV)=1−0.98=0.02 (2% of HIVIV-infected patients test negative)
● P(Ng∣N)=1−P(P∣N)=1−0.03=0.97 (97% of HIVIV-uninfected patients test negative)
Now, let's solve each part of the question:
a) What is the probability that a patient testing positive for HIVIV with this test
is infected with it?
We need to find P(HIV∣P). Using Bayes' theorem:
P(HIV∣P)=P(P)P(P∣HIV)⋅P(HIV)
To find P(P), the probability of testing positive, we can use the law of total probability:
P(P)=P(P∣HIV)⋅P(HIV)+P(P∣N)⋅P(N)P(P)=(0.98)(0.08)+(0.03)(0.92)P(P)=0.0784+0.0276
P(P)=0.106
Now we can find P(HIV∣P):
P(HIV∣P)=0.106(0.98)(0.08)P(HIV∣P)=0.1060.0784P(HIV∣P)≈0.7396
So, the probability that a patient testing positive for HIVIV is actually infected with it is
approximately 73.96%.
b) What is the probability that a patient testing positive for HIVIV with this test
is not infected with it?
We need to find P(N∣P). We can use Bayes' theorem again:
P(N∣P)=P(P)P(P∣N)⋅P(N)P(N∣P)=0.106(0.03)(0.92)P(N∣P)=0.1060.0276
P(N∣P)≈0.2604
Alternatively, since a patient is either infected or not infected,
P(N∣P)=1−P(HIV∣P)=1−0.7396=0.2604.
So, the probability that a patient testing positive for HIVIV is not infected with it is
approximately 26.04%.
c) What is the probability that a patient testing negative for HIVIV with this test
is infected with it?
We need to find P(HIV∣Ng). Using Bayes' theorem:
P(HIV∣Ng)=P(Ng)P(Ng∣HIV)⋅P(HIV)
To find P(Ng), the probability of testing negative, we can use the law of total
probability:
P(Ng)=P(Ng∣HIV)⋅P(HIV)+P(Ng∣N)⋅P(N)P(Ng)=(0.02)(0.08)+(0.97)(0.92)P(Ng)=0.0016+0.892
4
P(Ng)=0.894
Now we can find P(HIV∣Ng):
P(HIV∣Ng)=0.894(0.02)(0.08)P(HIV∣Ng)=0.8940.0016P(HIV∣Ng)≈0.00179
So, the probability that a patient testing negative for HIVIV is actually infected with it is
approximately 0.179%.
d) What is the probability that a patient testing negative for HIVIV with this test
is not infected with it?
We need to find P(N∣Ng). Using Bayes' theorem:
P(N∣Ng)=P(Ng)P(Ng∣N)⋅P(N)P(N∣Ng)=0.894(0.97)(0.92)P(N∣Ng)=0.8940.8924
P(N∣Ng)≈0.99821
Alternatively, since a patient is either infected or not infected,
P(N∣Ng)=1−P(HIV∣Ng)=1−0.00179=0.99821.
So, the probability that a patient testing negative for HIVIV is not infected with it is
approximately 99.821%.
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