Probability question involving conditional probability A medical patient is diagnosed with a condition that is fatal $60$% of the time. One possible treatment involves a surgical procedure. Research has shown that 40% of survivors had surgery and 10% of non-survivors had surgery. What is the probability of the patient surviving the condition if they have surgery? So I say: $P(L) = live$ $P(D) = die$ $P(S) = surgery$ $P(N) = no surgery$ We want: $P(L \mid S) = \frac{P(L \cap S)}{P(S)}$ I'm lost as to how we can write in terms of probability "40% of survivors had surgery and 10% of non-survivors had surgery" I'm hitting a wall which intuitively should be easy to pass but I can't seem to get my head around the problem. Anyone know how to approach this with a better intuition than me for probability?

You write that $P(S|L)=0.4$ (the probability that the person had surgery given that they lived is 0.4) and $P(S|D)=0.1$ (the probability that the person had surgery given that they died is 0.1). $$ P(L\cap S)=P(L|S)\times P(S)=P(S|L)\times P(L) $$ Thus $$ P(L|S)=P(S|L)\times{P(L)\over P(S)} $$ Now $$ \begin{align} P(S)&=P(S|L)P(L)+P(S|D)P(D)\\\ &=0.4\times 0.4+0.1\times0.6\\\ &=0.22 \end{align} $$ Can you finish it?