Having difficulty with the following homework assignment:
$ \begin{array}{c|c|c|c} \text{Test} & \text{Disease} & \text{No Disease} & \\ \hline \\ + & 436 & 5 & 441\\ - & 14 & 495 & 509\\ \hline \\ & 450 & 500 & 950 \end{array}$
What is the probability that a randomly selected individual will have an erroneous test result? That is, what is the probability that an individual has the disease and receives a negative test result ($-$) or the individual does not have the disease and receives a positive test result ($+$)?
I thought the answer should be:
$\begin{align} a &= P(-\text{ and }D) + P(+\text{ and }D)\\ & = P(-) \cdot P(D) + P(+)\cdot P(ND)\\ & = P(-)\cdot\frac{450}{950} + P(+)\cdot\frac{500}{950} \end{align}$
How do you calculate $P(-)$ and $P(+)$?
The answer is given as $0.010021$.
Any help is appreciated.
Thank you!