Is my probability calculation correct 
$20\%$ of a class earned an $A$ in the mid term exam. Of those $20\%$, $50\%$ of them also scored an $A$ in the final exam. Of the people who scored less than $A$ in the mid-term, $20\%$ of them scored an $A$ in the final. If I pick up a random final paper with a score of $A$, what is the chance that the student also scored an $A$ in the mid term.

Ok so my calculations are: $10\%$ of students scored $A$ in mid term and final. $16\%$ of students scored A in final only. So, the chance that a student with a final $A$ score, who also got $A$ in mid term, in both is simply $$\frac{10}{10+16}=\frac{10}{26}$$ approximately $38\%$.
I'd like to think I am right but it seems a bit too easy which leads me to think I am overlooking something. 
 A: Let's mind in this way: we define $T$ as the total amount of people who took the test.
Hence
$$20\% T = \frac{20}{100} T = A_{\text{mid term}} = \frac{1}{5} T$$
$$50\% (20\% T) = \frac{500}{10000} T = A_{\text{final}} = \frac{1}{10} T$$
$$20\% (80\% T) = \frac{1800}{10000} T = A_{\text{final}}^{\text{less than A in midterm}} = \frac{4}{25} T$$
Now the total amount of people who got $A$ in the final exam is
$$\frac{1}{10} T + \frac{4}{25} T = \frac{13}{50} T$$
To get the total amount of possible people who scored $A$ in both the test we just need to calculate
$$P = \frac{\frac{1}{10}}{\frac{13}{50}} = \frac{1}{10}\cdot \frac{50}{13} = \frac{5}{13}$$
Which is exactly your answer.
A: You are correct. Formally, you can define the events:


*

*$A_m:=$ a student scored $A$ in the mid-term exam.

*$A_f:=$ a student scored $A$ in the final exam.


with given probabilities $P(A_m)=0.2, P(A_f\mid A_m)=0.5, P(A_f \mid A_m^c)=0.2$ and you are looking for $P(A_m\mid A_f)$. By Bayes Rule (and the law of total probability for the denominator) you get that \begin{align}P(A_m \mid A_f)&=\frac{P(A_f\mid A_m)P(A_m)}{P(A_f)}=\frac{P(A_f\mid A_m)P(A_m)}{P(A_f\mid A_m)P(A_m)+P(A_f\mid A_m^c)P(A_m^c)}\\[0.2cm]&=\frac{0.5(0.2)}{0.5(0.2)+0.2(0.8)}=\frac{10}{10+16}=\frac{10}{26}\end{align}

The only difference in your approach is that you calculated $P(A_f\cap A_m)$ from the formula $$P(A_f\cap A_m)=P(A_f\mid A_m)P(A_m)=0.2(0.5)=0.10$$ or $10\%$ and similarly $P(A_f^c \cap A_m)$. These calculations gave you $P(A_f)$ from the formula $$P(A_f)=P(A_f\cap A_m)+P(A_f\cap A_m^c)$$ which is of course perfectly ok!
