Respuesta :
Answer:
0.8015 = 80.15% probability that the student has the flu
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Has high temperature.
Event B: Has the flu
Probability of a student having high temperatures:
90% of 35%(have the flu)
12% of 100 - 35 = 65%(do not have the flu). So
[tex]P(A) = 0.9*0.35 + 0.12*0.65 = 0.393[/tex]
Probability of having high temperatures and the fly?
90% of 35%, so
[tex]P(A \cap B) = 0.9*0.35 = 0.315[/tex]
If a student has a high temperature, what is the probability that the student has the flu?
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.315}{0.393} = 0.8015[/tex]
0.8015 = 80.15% probability that the student has the flu
