Pr(G∣E) : Probability of losing the first appeal but succeeding in the secondEvents:
Q=allowed in the country
A=minor mistake
B=lie
C=missing documents
D=successful first appeal
E=failed the first appeal
G= successful second appeal
What you have so far:
\(\displaystyle \Pr(Q)=\Pr(D)+\Pr(G|E)\)
\(\displaystyle \Pr(Q)=\Pr(D\cap A) + \Pr(D\cap B) + \Pr(D\cap C)+\Pr(G|E)\)
\(\displaystyle \Pr(Q)=\Pr(D|A)\Pr(A) + \Pr(D|B)\Pr(B) + \Pr(D| C)\Pr(C)+\blue{\Pr(G|E)}\)
Note that: [imath]1-\Pr(D)=\Pr(E)[/imath]
What do you think about the probability in the blue?
The probability of a successful second appeal is independent and the same as the first appeal because the conditional probabilities (0.95, 0.93, 0.93) remained the same. In other words, your chances aren't worse or better than the previous appeal. The second appeal didn't have any "memory" of the first appeal.Pr(G∣E) : Probability of losing the first appeal but succeeding in the second
This needs to contain P(G|A)*P(A)*P(E|A) =0.95*0.7*(1-0.95) , P(G|B)*P(B)*P(E|B) = (1-0.93)*0.1*0.93 , P(G|C)*P(C)*P(E|C) = (1-0.93)*0.2*0.93
Is right? I am not sure how to express it in a formula