Question

a canditate takes three tests in succession and the probability of passing the first test is p . the probability of passing each succeeding test is p or p/2 according as he passes or fails in the preceeding one. the candidate is selected if he passes atleast 2 tests. the probability that the candidate is selected is ?

Answer 1

Notice that if $p$ is the probability of pass, then the probability of fail is $1-p$.

The candidate can pass atleast 2 tests in 4 ways and the probabilities are given by :
1) passing all 3 tests : $p*p*p=p^3$
2) passing first two tests and failing third test : $p*p*(1-p)=p^2(1-p)$
3) passing first test, failing second and passing third : $p*(1-p)*p/2=p^2(1-p)/2$
4) failing first test, passing second and third : $(1-p)*p/2*p=p^2(1-p)/2$

Add them up to get the probability for candidate to be selected :
$p^3+p^2(1-p)+p^2(1-p)/2+p^2(1-p)/2=2p^2-p^3$

Answer 2

We need to know the probability that he/she passes at least two tests. 
Let the three tests be named as A, B and C (in order). The required probability is 
$P(A \cap B \cap C)+P(A' \cap B \cap C)+P(A \cap B' \cap C)+P(A \cap B \cap C') \\ =p.p.p + (1-p).\frac{p}{2}.p +p.(1-p).\frac{p}{2} + p.p.(1-p) \\ = p^{3}+2p^{2}-2p^{3} \\ =2p^{2}-p^{3}$ where we have used $P(R \cap S)=P(R).P(S|R)$.
I am not so sure about it, in any case feel free to report it.