Answer:
0.324
Explanation:
From the given information:
• The probability of exposure to Lyme's disease, P(L) = 0.01
,• The probability of testing positive given that they are exposed, P(+|L)=0.95
,• The probability of testing positive given that they are not exposed, P(+|L')=0.02
From the first probability:
[tex]\begin{gathered} P(L)+P(L^{^{\prime}})=1 \\ P(L^{^{\prime}})=1-P(L)=1-0.01=0.99 \end{gathered}[/tex]We want to find the probability that given that a randomly selected person tested positive, she has in fact been exposed to the disease, P(L|+).
By the Baye's theorem for conditional probability:
[tex]P(L|+)=\frac{P(+|L)P(L)}{P(+|L)P(L)+P(+|L^{\prime})P(L^{\prime})}[/tex]Substitute the values above:
[tex]\begin{gathered} P(L\lvert+)=\frac{0.95\times0.01}{0.95\times0.01+0.02\times0.99} \\ \approx0.324 \end{gathered}[/tex]The probability is 0.324.
