Conditional Probability

[393sh]

题：

 

[Deleted member 4993]

View attachment 31678
题：
View attachment 31679

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[393sh]

 

[BigBeachBanana]

Events:
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

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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?

 

[393sh]

Events:
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?

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

 

[BigBeachBanana]

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

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.
[imath]\Pr(Q)=\Pr(D)+\Pr(G\cap E)\\ \Pr(Q)=\Pr(D)+\Pr(G|E)\cdot \Pr(E)\\ \Pr(Q)=\Pr(D) +\Pr(D)\cdot \Pr(E)\\ \Pr(Q)=\Pr(D) +\Pr(D)[1-\Pr(D)] [/imath]