1. ## probability of illnes

A= the paitent is ill: P(A)=0,001
A*= the paitent is not ill
B= the diagnos shows ill
B*=the diagnos shows not ill

given: p(B|A)= 0,8
p(B*|A*)= 0,9

Question: What the probability that the paitent is ill and the diagnos shows ill?

started with: P(A|B)= P(A∩B)/P(B)= (P(A)xP(B|A))/P(B)

.....

2. Hello, sf1903!

$\displaystyle \begin{array}{ccc}A &= & \text{paitent is ill} \\ A^* & = & \text{patient is not ill} \\ B & = & \text{test shows ill} \\ B^* & = & \text{test shows not ill} \end{array} \qquad P(A) \:= \:0.001\;{\color{blue}[1]}$

Given: .$\displaystyle \begin{array}{cccc}P(B|A) &=& 0.8 & {\color{blue}[2]}\\ P(B^*|A^*) &= & 0.9\end{array}$

What the probability that the paitent is ill and the test shows ill?
They are asking for: .$\displaystyle P(A \wedge B)$

Using Bayes' Theorem on [2]: .$\displaystyle P(B|A) \:=\:\frac{P(A \wedge B)}{P(A)} \:=\:0.8$

Substitute [1]: .$\displaystyle \frac{P(A \wedge B)}{0.001} \:=\:0.8$

Therefore: .$\displaystyle P(A \wedge B) \;=\;(0.001)(0.8) \;=\;0.0008$

3. Originally Posted by sf1903
A= the paitent is ill: P(A)=0,001
A*= the paitent is not ill
B= the diagnos shows ill
B*=the diagnos shows not ill

given: p(B|A)= 0,8
p(B*|A*)= 0,9

Question: What the probability that the paitent is ill and the diagnos shows ill?

started with: P(A|B)= P(A∩B)/P(B)= (P(A)xP(B|A))/P(B)

.....
b) how many % of the patients are diagnosed wrong?

P(B*
|S)=0,2
P(B
| S*)=0,1

???

4. Originally Posted by sf1903
b) how many % of the patients are diagnosed wrong?

P(B*|S)=0,2
P(B| S*)=0,1

???
Using the notation you introduced in your first post (what's S and S* meant to be?): Pr(B | A') = 0.1 and Pr(B' | A) = 0.2.

So the answer is 0.1 + 0.2 = 0.3