Probability of Alzheimers Disease

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!

• I knew the answer, but forgot it. :-) – Asaf Karagila Jan 28 '12 at 10:00

The total number of individuals is $N=950$. A test may be erroneous, either because the patient has no disease and the test is positive, and there are $N_{+,ND}=5$ such individuals, or because the patient has the disease and the test is negative, and there are $N_{-,D}=14$ such individuals. The total number of individuals with an erroneous test is $N_e=N_{+,ND}+N_{-,D}=5+14=19$, hence the probability that a randomly selected individual will have an erroneous test result is $$N_e/N=19/950=0.02.$$ Your solution uses $P(-\,\text{and}\,D)=P(-)P(D)$ and $P(+\,\text{and}\,ND)=P(+)P(ND)$. These identities are not true here. They would hold if and only if the result to the test and the health of the individuals were independent.
The rational number one can think about, which is closest to the result $0.010021$ which you mention, seems to be $5/499=0.01002004$, but even this number does not round up to $0.010021$, and, anyway, I have no idea what erroneous reasoning could produce this answer from the data $436$, $5$, $14$ and $495$.
I've played with the numbers and noticed that the total number of patients getting erroneous results is $19\cdot (14+5)$ and $950$ divided by $19$ is $50$. I have no idea what to do with that, maybe you or someone else will have, or it may be only a coincidence!