conditional probability of a union A patient would like to take a test to determine if he has a nasty disease. Let the variable A denote that
the patient has the disease and the variable B denote a positive test. The following assumptions apply:
• The probability that the test is positive given the patient has the disease is 99%.
• The probability that the test is positive given the patient does not have the disease is 5%.
• The rate of occurrence of the disease in the general population is 0.1%.
Consider a second test for the same disease. A positive result for this test is denoted by C. The following
assumptions apply:
• The probability that the test is positive given the patient has the disease is 80%.
• The probability that the test is positive given the patient does not have the disease is 0.01%.
Now that both tests have been taken, assume conditional independence of these tests.
Both tests give a positive answer. What is the probability that the patient has the disease?
so my question is do i apply P(A|BnC) ?if so how do i deduce this
 A: First, let us summarise the information we are given:-
$$P(A)=0.001\Rightarrow P(\overline{A})=1-P(A)=0.999$$
For the first test:-
$$P(B|A)=0.99$$
$$P(B|\overline{A})=0.05$$
For the second test:-
$$P(C|A)=0.8$$
$$P(C|\overline{A})=0.0001$$
In order to calculate $P(A|B,C)$, let us invoke Bayes' Rule:-
$$P(A|B,C)=\frac{P(A,B,C)}{P(B,C)}=\frac{P(B,C|A)P(A)}{P(B,C|A)P(A)+P(B,C|\overline{A})P(\overline{A})}$$
We are told that the tests are conditionally independent given the disease status of the patient (highlighted in blue, below), so that
$$\begin{align}
P(A|B,C)&=\frac{\color{blue}{P(B|A)P(C|A)}P(A)}{\color{blue}{P(B|A)P(C|A)}P(A)+\color{blue}{P(B|\overline{A})P(C|\overline{A}})P(\overline{A})}\\&=\frac{0.99\times0.8\times0.001}{(0.99\times0.8\times0.001)+(0.05\times0.0001\times0.999)}\\&=0.99373\approx99.4\%\end{align}$$
A: You don't want $P(A | B \cup C)$, you want $P(A | B \cap C)$. Here are some hints:
By the definition of conditional probability, and then by independence:
$$P(A | B \cap C) = \frac{P(A \cap B \cap C)}{P(B \cap C)} = \frac{P(A \cap B \cap C)}{P(B)P(C)}.$$
The numbers that you have are
$P(A) = .001$, $P(B |A) = 0.99$, $P(B | \overline{A}) = .05$, $P(C|A) = 0.8$, $P(C|\overline{A}) = 0.0001$.
Use the definitions of conditional probability. Try to take it from here.
