How to calculate conditional probability with limited information? Recently I am doing a question

A new screening test (Test $A$) is now purposed. It's known that $23\%$ of patients who are $A$ positive will have disease $D$. It's known that $10\%$ of the population are detected to be $A$ positive. The probability that people have disease $D$ is $5\%$. Show that the probability of disease $D$ given $A$ negative is $3.00\%$.

I tried to draw a tree diagram

However, when the probability of $P(A-\mid D)\times P(D)$ is unknown, how can I prove it?
Thank you!!
 A: There are (at least) a couple of ways to go about this.  One is to draw up a table, imagining you have a perfectly representative group of $1000$ people:
$$
\begin{array}{|c|c|c|c|} \hline
& \text{A negative} & \text{A positive} & \text{A sum} \\ \hline
\text{D negative} &  &  & 950 \\ \hline
\text{D positive} &  & 23 & 50 \\ \hline
\text{D sum} & 900 & 100 & 1000 \\ \hline
\end{array}
$$
Note that:

*

*The first column must sum to $900$, because $9/10$ of the people test negative; the second column sums to $100$, because $1/10$ of the people test positive.


*Conversely, the first row must sum to $950$, because $19/20$ of the people don't have the disease; the second row must sum to $50$, because $1/20$ of the people have the disease.


*Finally, we know that of the $100$ people who test positive, $23$ of them actually have the disease.
If you fill in the remainder of the table, you should have your answer.

Another way is to use Bayes's theorem, along with a little logic.  We know that:

*

*$P(\text{A positive}) = 1/10$

*$P(\text{D positive}) = 1/20$

*$P(\text{D positive} \mid \text{A positive}) = 23/100$
If you note that
$$
P(\text{A positive} \mid \text{D positive})
    = \frac{P(\text{D positive} \mid \text{A positive}) P(\text{A positive})}
           {P(\text{D positive})}
$$
and
$$
P(\text{A negative} \mid \text{D positive})
    = 1-P(\text{A positive} \mid \text{D positive})
$$
then you can find
$$
P(\text{D positive} \mid \text{A negative})
    = \frac{P(\text{A negative} \mid \text{D positive}) P(\text{D positive})}
           {P(\text{A negative})}
$$
A: I use the following notation:

*

*event $A$: Test is positive $\quad$ ($\overline{A}$: Test is negative)

*event $D$: Patient have disease

Variant 1
If you know Bayes' theorem, then use it for the first time to get
$$P(D|\overline{A})=P(\overline{A}|D)\cdot\frac{P(D)}{P(\overline{A})}=[1-P(A|D)]\cdot \frac{P(D)}{P(\overline{A})}$$
Now apply Bayes' theorem for the second time on $P(A|D)$ to get
$$P(D|\overline{A})=\Big[1-\frac{P(D|A)P(A)}{P(D)}\Big]\cdot\frac{P(D)}{P(\overline{A})}$$
and this will end up in
$$P(D|\overline{A})=\Big[1-\frac{23\% \cdot 10\%}{5\%}\Big]\cdot\frac{5\%}{90\%}=3\%$$
Variant 2
If you don't know Bayes' theorem, then use a CORRECT tree diagram

I hope this will help you:
\begin{align}
5\%=P(D)&=P(A\cap D)+P(\overline{A}\cap D)\\
&=P(A)\cdot P(D|A)+P(\overline{A})\cdot P(D|\overline{A})\\
&=10\%\cdot 23\%+90\%\cdot P(D|\overline{A})
\end{align}
Rearranging will lead to
$$P(D|\overline{A})=3\%$$
A: Given that,
P(D|A) = o.23, P(A) = 0.10, P(D) = 0.05
Now, P(D|A) = P(D ∩ A) / P(A) => P(D∩A) = 0.023
Hence, P(D|A-) = P(D ∩ A-) / P(A-) = 0.03
Hope that this will solve your problem :)
