Answer with explanation:
Probability that the student is suffering from lice the test shows Positive
[tex]=P(\frac{PT}{L})=0.2632[/tex]
Probability that the student is not suffering from lice and the test shows Positive
[tex]=P(\frac{PT}{N L})=0.0432[/tex]
Abbreviation used
L = Student has lice
N L=Student has no lice
P T=Test shows Positive
Probability that a student has lice, given that the student tested positive
[tex]P(\frac{L}{P})=\frac{P(\frac{PT}{L})}{P(\frac{PT}{L})+P(\frac{PT}{NL})}\\\\P(\frac{L}{P})=\frac{0.2632}{0.2632 +0.0432}\\\\P(\frac{L}{P})=\frac{0.2632}{0.3064}\\\\P(\frac{L}{P})=0.8590[/tex]
In terms of Percentages Required Probability
= 0.8590 × 100
= 85.90 %
Option C
