Alice tries for 4 job interviews A,B,C,D with the probabilities to pass the interviews
$Pr[A]=0.1$
$Pr[B]=0.2$
$Pr[C]=0.3$
$Pr[D]=0.4$
There are no dependence to get accepted for different interviews.
Alice will try the 4 interviews inorder and will stop when she succeeds in two.
If Alice has succeeded in interview C ,what is the probability for Alice to do the D interview?
I thought of using Bayes rule for two events $Pr[D|C]=\frac{Pr[C|D]\cdot Pr[D]}{Pr[C|D]\cdot Pr[D] +Pr[C|\bar{D}]\cdot Pr[\bar{D}]}.$
I thought cause the events are independent i can calculate :$Pr[C|D]=\frac{Pr[C\cap{D}]}{Pr[D]}=\frac{1-(Pr[A\cap{B}])+Pr[A\cap{C}]+Pr[B\cap{C}])}{Pr[D]}=\frac{1-0.11}{0.4}$.
But i get a wrong probability value.