Given the probabilities of a false positive and a false negative, find the probabilities of true positive and true negative If I am given that out of $1000$ individuals where $60$ use a drug. I am given the probability of a false positive is $.009$ and the probability of a false negative is $.10$. I'm trying to find the true positive and true negative. 
Let $D$ be the even that the user uses a drug and $D^C$ the event the user is not a drug user. 
I know that $P(D)=.06$ and $P(D^C)=.994$ as well as $P(+|D^c)=.009$ and $P(-|D)=.10$. So that means that $P(+|D^C)=.009\implies .009*990=8.91$ people. Also $P(-|D)=.10\implies 60*.10=6$ people. Does it follow that $P(-|D^C)=981.09/990=.991$ and $P(+|D)=54/60=.90$?. 
 A: You computed too much, but the numbers you got are correct.
With no computation, $\Pr(-|D^c)=1-0.009=0.991$ and $\Pr(+|D)=1-0.10=0.9$. 
Some steps in your computation were technically not right. For example, it is not true that $\Pr(-|D)=6$ (people). Probabilities are always between $0$ and $1$. 
Now you need to solve the real problem, which presumably has to do with finding $\Pr(D|+)$. For this you will need the definition of conditional probability, or Bayes' Formula. 
A: First, $60$ drug users out of $1000$ is $\Pr[D] = 0.06$, not $0.006$.
The probability of a false positive is $\Pr[+ \mid \bar D] = 0.009$, where $\bar D = D^c$ in your notation.  Similarly, $\Pr[- \mid D] = 0.10$.  By Bayes' theorem, we then have $$\Pr[D \mid +] = \frac{\Pr[+ \mid D]\Pr[D]}{\Pr[+]} = \frac{(1 - \Pr[- \mid D])\Pr[D]}{\Pr[+]}.$$  We know all the RHS probabilities except $\Pr[+]$.  So we use the law of total expectation:  $$\Pr[+] = \Pr[+ \mid D]\Pr[D] + \Pr[+ \mid \bar D]\Pr[\bar D].$$  Again, since $\Pr[+ \mid D] = 1 - \Pr[- \mid D]$, all we need to do is substitute and you will get your answer.
Another method is to construct a $2 \times 2$ table:  from the given information, you can directly fill in the following cell frequencies:  $$\begin{array}{c|c c|c} & + & - & \\ \hline D & ? & 6 & 60 \\ \bar D & 8.46 & ? & 940 \\ \hline & ? & ? & 1000 \end{array}$$  Of course, actual counts must be nonnegative integers, but for the purposes of computing probabilities or proportions, this is fine.  We can then fill in the rest of the empty spaces:  $$\begin{array}{c|c c|c} & + & - & \\ \hline D & 54 & 6 & 60 \\ \bar D & 8.46 & 931.54 & 940 \\ \hline & 62.46 & 937.54 & 1000 \end{array}$$  Now it is trivial to compute the desired probability of a true positive:  $$\Pr[D \mid +] = 54/62.46 = 0.864553.$$
A: 
If I am given that out of $1000$ individuals where $60$ use a drug. I am given the probability of a false positive is $.009$ and the probability of a false negative is $.10$. I'm trying to find the true positive and true negative. 
Let $D$ be the even that the user uses a drug and $D^C$ the event the user is not a drug user. 

We know $P(D)=0.06, P(D^C)=0.94, P(+|D^c)=0.009, P(-|D)=0.10$
Just use total probability:
$\begin{align}
P(+\mid D) & = 1 - P(-\mid D)  & \text{True Positive}
\\ & = 0.90
\\[1ex]
 P(-\mid D^c) & = 1-P(+\mid D^c)  & \text{True Negative}
\\ & = 0.001
\\[2ex]\text{Then we can find:}
\\
P(D\mid +) & = \frac{P(+\mid D)P(D)}{P(+\mid D)P(D)+P(+\mid D^c)P(D^c)} & \text{Positive Confidence}
\\ & = \frac{0.90\cdot 0.06}{0.90\cdot 0.06+0.009\cdot 0.94}
\\ & \approx 0.86[5]
\end{align}$
