Answer:
The percentage of people having cholera is 89.19%.
Step-by-step explanation:
According to the Bayes' theorem the total probability of A is:
[tex]P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})[/tex]
Let X = a person has chorea and Y = the test is positive.
Given:
[tex]P(X^{c}|Y)=0.05\\P(X|Y^{c})=0.12\\P(Y)=0.93[/tex]
The value of [tex]P(X|Y)[/tex] is:
[tex]P(X|Y)=1-P(X^{c}|Y)=1-0.05=0.95[/tex]
Compute the value of P (X) as follows:
[tex]P(X)=P(X|Y)P(Y)+P(X|Y^{c})P(Y^{c})\\=(0.95\times0.93)+(0.12\times(1-0.93))\\=0.8835+0.0084\\=0.8919[/tex]
The percentage of people suffering form cholera is, 0.8919 × 100 = 89.19%.
Thus, the percentage of people having cholera is 89.19%.