MHB How will the doctor decide to operate on Mr. Peter?

[mathmari]

Hey!

We assume that for a particular illness, a doctor recommends a dangerous surgery if, after a clinical examination and by laboratory tests, he is 80% sure that his patient suffers from it, or, in other case, he recommends further costly examinations. Laboratory tests make a good diagnosis in 99% of cases for non-diabetics and in 70% of cases for diabetics.

After a clinical examination, the doctor is 60% sure that Mr Peter suffers from the disease. Laboratory tests that have a positive result (for the disease) are also done.
Will the doctor operate Mr. Peter, believing that Mr. Peter is not diabetic, or will he recommend further tests?
What will the doctor do if Mr. Peter after the results remembers that he is diabetic? Could you give me a hint how we could check that?

Let $x$ be the probability that the patient suffers from the disease.

In the first case we consider that Mr Peter is not diabetic. Then the probability that he is ill, believing that he is non-diabetic is equal to $0.99\cdot x+0.01\cdot (1-x)$ and this must be equal to the estimation of the docrot, i.e. $0.99\cdot x+0.01\cdot (1-x)=0.6$.

Is this correct?

At the case where Mr Peter remembers that he is diabetic, do we have that $0.7\cdot x+0.3\cdot (1-x)=0.6$ ? Or have I understood wrong the exercise statement? (Wondering)

 

[I like Serena]

Hey!

We assume that for a particular illness, a doctor recommends a dangerous surgery if, after a clinical examination and by laboratory tests, he is 80% sure that his patient suffers from it, or, in other case, he recommends further costly examinations. Laboratory tests make a good diagnosis in 99% of cases for non-diabetics and in 70% of cases for diabetics.

After a clinical examination, the doctor is 60% sure that Mr Peter suffers from the disease. Laboratory tests that have a positive result (for the disease) are also done.
Will the doctor operate Mr. Peter, believing that Mr. Peter is not diabetic, or will he recommend further tests?
What will the doctor do if Mr. Peter after the results remembers that he is diabetic? Could you give me a hint how we could check that?

Let $x$ be the probability that the patient suffers from the disease.

In the first case we consider that Mr Peter is not diabetic. Then the probability that he is ill, believing that he is non-diabetic is equal to $0.99\cdot x+0.01\cdot (1-x)$ and this must be equal to the estimation of the docrot, i.e. $0.99\cdot x+0.01\cdot (1-x)=0.6$.

Is this correct?

At the case where Mr Peter remembers that he is diabetic, do we have that $0.7\cdot x+0.3\cdot (1-x)=0.6$ ? Or have I understood wrong the exercise statement? (Wondering)

Hey mathmari! (Smile)

I read it as:
P(Ill | positive clinical test) = 60%
P(Ill | positive laboratory test AND non-diabetic) = 99%

So:
P(¬Ill | positive clinical test) = 40%
P(¬Ill | positive laboratory test AND non-diabetic) = 1%

Assuming independence of the clinical test and the laboratory test, I think that means:
P(¬Ill | positive clinical test AND positive laboratory test AND non-diabetic) = 40% x 1% = 0.4%

Thus:
P(Ill | positive clinical test AND positive laboratory test AND non-diabetic) = 99.6% > 80%And if Mr. Peter is diabetic, this becomes:
P(Ill | positive clinical test AND positive laboratory test AND diabetic) = 1 - 40% x 30% = 88% > 80%
(Thinking)

 

[mathmari]

I read it as:
P(Ill | positive clinical test) = 60%
P(Ill | positive laboratory test AND non-diabetic) = 99%

So:
P(¬Ill | positive clinical test) = 40%
P(¬Ill | positive laboratory test AND non-diabetic) = 1%

Assuming independence of the clinical test and the laboratory test, I think that means:
P(¬Ill | positive clinical test AND positive laboratory test AND non-diabetic) = 40% x 1% = 0.4%

Thus:
P(Ill | positive clinical test AND positive laboratory test AND non-diabetic) = 99.6% > 80%And if Mr. Peter is diabetic, this becomes:
P(Ill | positive clinical test AND positive laboratory test AND diabetic) = 1 - 40% x 30% = 88% > 80%
(Thinking)

So, we have that \begin{align*}P&(\text{Ill | positive clinical test AND positive laboratory test AND non-diabetic}) \\ &= 1 - \text{P(¬Ill | positive clinical test AND positive laboratory test AND non-diabetic) } \\ & = 1 - \text{P(¬Ill | positive clinical test)*P(¬Ill | positive laboratory test AND non-diabetic)}\end{align*} right? Why do we not take \begin{align*}P&(\text{Ill | positive clinical test AND positive laboratory test AND non-diabetic}) \\ & = \text{P(Ill | positive clinical test)*P(Ill | positive laboratory test AND non-diabetic)}\end{align*} ? Why is this wrong?

(Wondering)

 

[I like Serena]

Actually, I think we shouldn't calculate it in either way.
We can't treat the conditional part that way.

Let's look at it differently.

First off, given that Mr. Peter is a non-diabetic, we have a good diagnostic result of the laboratory test of 99%.
I'm interpreting it to mean that sensitivity and specificity of the test are both 99%.
That is:
$$P(\text{Positive laboratory test} \mid Ill) = P(\lnot\text{Positive laboratory test} \mid \lnot Ill) = 99\% \tag 1$$

Now, based on the clinical test, we know that Mr. Peter is in a population where 60% has the illness.
That is:
$$P(Ill) = 60\%\tag 2$$

So within the given population (positive clinical test and non-diabetics), and assuming independence of the clinical test and the laboratory test, we want to know:
$$P(Ill \mid +) = P(Ill \mid \text{Positive laboratory test}) = \quad ?$$

From Bayes' theorem we get:
$$\begin{aligned}P(Ill\mid +) &= \frac{P(+\mid Ill) P(Ill)}{P(+)}
= \frac{P(+\mid Ill) P(Ill)}{P(+\mid Ill) P(Ill)+P(+\mid \lnot Ill) P(\lnot Ill)} \\
&= \frac{P(+\mid Ill) P(Ill)}{P(+\mid Ill) P(Ill)+\Big(1 - P(-\mid \lnot Ill)\Big)\Big(1 - P(Ill)\Big)} \\
&= \frac{0.99 \times 0.60}{0.99 \times 0.60 +(1 - 0.99) \times (1 - 0.60)} \\
&= 99\% > 80\%
\end{aligned}\tag 3$$
Thus the doctor will recommend to operate on Mr. Peter.

Redoing this assuming that Mr. Peter is diabetic, we get:
$$\begin{aligned}P(Ill\mid +)
&= \frac{0.70 \times 0.60}{0.70 \times 0.60 +(1 - 0.70) \times (1 - 0.60)} \\
&= 78\% < 80\%
\end{aligned}\tag 4$$
In other words, the doctor will recommend further costly examinations.

 

[mathmari]

I understand! Thank you so much! (Sun)