# Thread: Probability that someone has the disease

1. Hey!! The percentage of people that have a disease A is $0,01$.
We apply twice a test for that disease, each of which give the correct answer with probability $0,95$.
What is the probability that someone has that disease if at least one test is positive and what is the probability if both tests are positive?

I have done the following:

We have that $$P(\text{at least one positive})=1-P(\text{no positive})=1-P(NN)$$ where N: "negative".

Let D be the event "has the disease".

Then we have that $$P(NN)=P(NN\cap D)+P(NN\cap D^C)=P(NN\mid D)\cdot P(D)+P(NN\mid D^C)\cdot P(D^C)$$ We have that $P(D)=0,01$ and $P(D^C)=0,99$.

Does it hold that $P(NN\mid D)=P(N\mid D)^2$ and $P(NN\mid D^C)=P(N\mid D^C)^2$ ?

If yes, does it hold that $P(N\mid D)=0,05$ and $P(N\mid D^C)=0,95$ ?

That would mean that we get $$P(NN)=0,05^2\cdot 001+0,95^2\cdot 0,99$$ Is this correct?   Reply With Quote

2.

3. Originally Posted by mathmari Does it hold that $P(NN\mid D)=P(N\mid D)^2$ and $P(NN\mid D^C)=P(N\mid D^C)^2$ ?
If the two tests are independent, yes. Quote:
If yes, does it hold that $P(N\mid D)=0,05$ and $P(N\mid D^C)=0,95$ ?
Yes. Quote:
That would mean that we get $$P(NN)=0,05^2\cdot 001+0,95^2\cdot 0,99$$ Is this correct? Except for the typo where you meant "0.01" and not "0.001", yes.

You know you have not yet answered either question, right?  Reply With Quote Originally Posted by tkhunny Except for the typo where you meant "0.01" and not "0.001", yes.
Ah ok!

So, we have that $$P(NN)=0.05^2\cdot 0.01+0.95^2\cdot 0.99=0.8935$$ Originally Posted by tkhunny You know you have not yet answered either question, right?
For the first question we have the following:
$$P(\text{at least one positive})=1-P(\text{no positive})=1-P(NN)=1-0.8935=0.1065$$
Is this correct?

At the suggested solution the result is $0, 093661972$. Can that be? For the second question we have the following:
\begin{align*}P(PP)&=P(PP\cap D)+P(PP\cap D^C) \\ & =P(PP\mid D)\cdot P(D)+P(PP\mid D^C)\cdot P(D^C)\\ & =P(P\mid D)^2\cdot P(D)+P(P\mid D^C)^2\cdot P(D^C)\\ & =0.95^2\cdot 0.01+0.05^2\cdot 0.99 \\ & =0.0115\end{align*}

Again the suggested result is different. It is $0, 784782609$.   Reply With Quote

5.  Reply With Quote Originally Posted by I like Serena Isn't the question for P(D|at least one positive) respectively P(D|both positive)? Ah yes!

So, we have the following: $$P(D|\text{at least one positive})=\frac{P(D\cap \text{at least one positive})}{P(\text{at least one positive})}=\frac{P( \text{at least one positive}\mid D)\cdot P(D)}{0.1065}=\frac{P( \text{at least one positive}\mid D)\cdot 0.01}{0.1065}$$ Is everything correct so far? To what is $P( \text{at least one positive}\mid D)$ equal?   Reply With Quote

7.  Reply With Quote Originally Posted by I like Serena Yep! And P(at least one positive|D)=1-P(NN|D).
I see!! Thank you!!   Reply With Quote

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