Finding probability for general cases For a student to qualify, he must pass at least two out of three exams. The probability that he will pass the 1st exam is $p$. If he fails in one of the exams then the probability of passing in the next exam is $p/2$ otherwise it remains the same. Find the probability that he will qualify.
My textbook answer reads  $2p^2 – p^3$.
This is possible if only the below cases are considered:


*

*He passes first and second exam.

*He passes first, fails in second but passes third exam.

*He fails in first, passes second and third exam.


But I think this is wrong since at least two out of three exams means,passing in first, second  and third exam is inclusive.  Someone please solve this paradox.
 A: See the probability tree diagram:

Adding the qualifying probabilities we get:
$$p^3+p^2(1-p)+p(1-p)\cdot \frac p2+(1-p)\cdot \frac p2 \cdot p=2p^2-p^3.$$
A: I get:
p^2 + (1-p)(p/2)p + p(1-p)(p/2)
Which is (pass pass either) + (fail pass pass) + (pass fail pass), and as you can see no cases are double counted here. If you then expand out the (1-p) terms and collect it together then it's the same as the textbook answer.
A: PP has probability $p^2$.
PFP has probability $p(1-p)\frac12p$.
FPP has probability $(1-p)\frac12pp$
Summation results in a probability of $2p^2-p^3$ to qualify.
A: All three passed: $p^3$.
First and second passed, third failed: $p^2(1-p)$.
First passed, second failed, third passed: $p(1-p)p/2$.
First failed, next two passed: $(1-p)(p/2)^2$.
Then add all four probabilities. The answer would be $7p^2/4-3p^3/4$.
UPDATE. I understood it so that if once failed, the probability of success remains at most $p/2$. But I suppose this is not correct. So the case "First failed, next two passed" leads to $(1-p)(p/2)p$ and the whole probability is indeed $2p^2-p^3$. The problem is badly stated, by the way.
A: Let me denote by $E_i$ the event "Exam $i$ has been passed" and let $\overline{E}_i$ be its complement.
Then, you know that
$P(E_1)=p$, $P(E_2| \overline{E}_1 ) 
= P(E_3| \overline{E}_1 \cap \overline{E}_2 )
= P(E_3| \overline{E}_1 \cap E_2 )
= P(E_3| E_1 \cap \overline{E}_2 )
= p/2$ and
$P(E_2| E_1 ) = P(E_3| E_1 \cap E_2 ) = p$.
By $\sigma$-addivity, you want to calculate the probability 
$$
A=P(E_1\cap E_2 \cap E_3) 
+ P(E_1\cap E_2\cap \overline{E}_3) 
+ P(E_1\cap \overline{E}_2 \cap E_3) 
+ P(\overline{E}_1 \cap E_2\cap E_3).$$
Let's calculate each term:
$$
P(E_1\cap E_2 \cap E_3)
= P(E_3 | E_1\cap E_2) P(E_1\cap E_2)
= p P( E_2 | E_1) P(E_1)
= p^3
$$
$$
P(E_1\cap E_2\cap \overline{E}_3)
= P(\overline{E}_3 | E_1\cap E_2) P( E_2 | E_1) P(E_1)
= (1-p) p^2
$$
$$
P(E_1\cap \overline{E}_2 \cap E_3) 
= P(E_3 | E_1\cap \overline{E}_2) P(E_1\cap \overline{E}_2)
= \frac{p}{2} P(\overline{E}_2 | E_1) P(E_1)
= \frac{p}{2} (1-p) p
$$
$$
P(\overline{E}_1 \cap E_2\cap E_3)
= P(E_3 | \overline{E}_1 \cap E_2 ) P (\overline{E}_1 \cap E_2)
= \frac{p}{2} P(E_2|\overline{E}_1) P(\overline{E}_1)
= \frac{p^2}{4}(1-p)
$$
To sum up, we obtain
$$
A =  p^3 + (1-p) p^2 + \frac{p}{2} (1-p) p + \frac{p^2}{4}(1-p)
%  =  p^3 + p^2- p^3  + \frac{p^2-p^3}{2}  + \frac{p^2}{4} - \frac{p^3}{4}
%  =  p^2   + 3 p^2 \frac{1-p}{4}  
%  =  p^2   +  \frac{3 p^2}{4}  - \frac{3 p^3}{4}  
 = \frac{7 }{4}p^2   - \frac{3 }{4}  p^3
$$
