$\begin{array}{1 1} 0.1654 \\ 0.1982 \\ 0.1827 \\ 0.1346 \end{array} $

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- Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: \(\;P(E_i/A)\)=\(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}\)

Let $E_1$: event that the person has disease $\rightarrow$ P($E_1$) = 0.1% = 0.0001.

Let $E_2$: event that the person does not have disease $\rightarrow$ P($E_2$) = 1 - 0.1% = 1 - 0.0001 = 0.999.

Let A: be the event that the test result is positive:

Probability that the result is positive given the person has disease = P(A|$E_1$) = 99% = 0.99

Probability that the result is positive given the person does not have disease = P(A|$E_2$) = 0.5% = 0.0005.

Since the two events are mutually exclusive and exhaustive, we can use Baye's theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)+P(A|E_2)}$

Using Baye's theorem, the probability that the person has disease given that his result is positive = P ($E_1$|A) =$\;\large \frac{0.0001 \;\times\; 0.99}{0.001 \;\times\; 0.99 + 0.9999\; \times\; 0.0005}$ = $\;\large\frac{0.00099}{0.005985} $= 0.1654.

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