How to calculate conditional probability with limited information?

Question

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!!

Answer 1

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.

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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

$$ P(\text{D positive} \mid \text{A negative}) = \frac{P(\text{A negative} \mid \text{D positive}) P(\text{D positive})} {P(\text{A negative})} $$

Answer 2

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 :)

Answer 3

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 $$P(D|\overline{A})=3\%$$