Answer:
1+1=2
50+4=54
54+2=56
Step-by-step explanation:
Using conditional probability, it is found that there is a 0.4235 = 42.35% probability that the patient really is HIV positive.
Conditional Probability
In which
- P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
- P(A) is the probability of A happening
In this problem:
- Event A: positive test.
- Event B: HIV positive.
The percentages involving a positive test are:
- 95% of 3%(positive)
- 4% of 100 - 3 = 97%(not positive).
Hence:
![P(A) = 0.95(0.03) + 0.04(0.97) = 0.0673](https://tex.z-dn.net/?f=P%28A%29%20%3D%200.95%280.03%29%20%2B%200.04%280.97%29%20%3D%200.0673)
The probability of both having a positive test and being HIV positive is:
![P(A \cap B) = 0.95(0.03)](https://tex.z-dn.net/?f=P%28A%20%5Ccap%20B%29%20%3D%200.95%280.03%29)
Then, the conditional probability is:
![P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.95(0.03)}{0.0673} = 0.4235](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D%20%3D%20%5Cfrac%7B0.95%280.03%29%7D%7B0.0673%7D%20%3D%200.4235)
0.4235 = 42.35% probability that the patient really is HIV positive.
A similar problem is given at brainly.com/question/14398287
Answer:
15/56 a²
Step-by-step explanation:
because we are ✖️ it