Let the Probability Begin




\begin{aligned} A &= \text{has the disease}, \qquad B = \text{tests positive} \\[14pt] P(A\mid B) &= \frac{P(A)P(B\mid A)}{P(A)P(B\mid A)+P(A^c)P(B\mid A^c)} \\[16pt] &= \frac{(0.001)(0.95)}{(0.001)(0.95)+(0.999)(0.005)} \\[16pt] &\approx 0.1598 \end{aligned}

 

 

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