In a medical survey conducted by a hospital, it was found that, of all the cancer deaths $20\%$ were due to oral cancer and $30\%$ due to lung cancer. It was also found that 60% of all cancer ailments result in death and 40% of all cancer patients have oral cancer. Find the probability that
1)a patient with oral cancer dies,
2)a cancer patient has lung cancer and he dies. Identify the sample space and all relevant events.
I assumed Sample space $S$ consisting of all cancer patients. Let $A_1$ be the event that a person has oral cancer, $A_2$ be the event that person has lung cancer and $A_3$ to be the event that person has another type of cancer. I per my interpretation the given data is $P(A_1)|B)=0.20, P(A_2|B)=0.30, P(A_3|B)=0.50, P(B)=0.60, P(A_1)=0.40$ and I need to find $P(B|A_1)$, which I am not able to find.
Hint:
Conditional probability gives you $P(A_1,B)=P(A_1 \mid B)P(B)=P(B \mid A_1)P(A_1)$
and so Bayes' theorem $$P(B \mid A_1) = \dfrac{P(A_1 \mid B)P(B)}{P(A_1)}$$