
MHB Master
#1
March 1st, 2018,
19:30
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 $$P(\text{at least one positive})=1P(\text{no positive})=1P(NN)$$ 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?

March 1st, 2018 19:30
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#2
March 2nd, 2018,
02:12
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?

MHB Master
#3
March 2nd, 2018,
04:17
Thread Author
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:
$$P(\text{at least one positive})=1P(\text{no positive})=1P(NN)=10.8935=0.1065$$
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 $0, 784782609$.

MHB Seeker
#4
March 2nd, 2018,
04:28
Isn't the question for P(Dat least one positive) respectively P(Dboth positive)?

MHB Master
#5
March 2nd, 2018,
04:42
Thread Author
Originally Posted by
I like Serena
Isn't the question for P(Dat least one positive) respectively P(Dboth 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?

MHB Seeker
#6
March 2nd, 2018,
04:47
Yep! And P(at least one positiveD)=1P(NND).

MHB Master
#7
March 2nd, 2018,
05:16
Thread Author
Originally Posted by
I like Serena
Yep! And P(at least one positiveD)=1P(NND).
I see!! Thank you!!