Confusion between conditional and joint probability This question might seem very trivial but I want to understand it in an intuitive way.
Let's say I have three options in a question and one of then is correct. I am randomly guessing the answer.
Now if I give the wrong answer in the first attempt I get a second chance
Then what is the probability that I will answer correctly?
I know that this answer should be P(first attempt correct) + P(second attempt correct)
Now what should be expression of P(second attempt correct)? It should be P(correct answer and first attempt wrong). But I am not understanding why it should not be P(correct answer|first attempt wrong)?
Can someone explain it in some intuitive way? And also what is the sample space and what are the correct events?
 A: 
I know that this answer should be P(first attempt correct) + P(second attempt correct) Now what should be expression of P(second attempt correct)? It should be P(correct answer and first attempt wrong). But I am not understanding why it should not be P(correct answer|first attempt wrong)? Can someone explain it in some intuitive way? And also what is the sample space and what are the correct events?

You only get the second attempt if the first fails. (This is why the events are disjoint.)
Therefore the answer should be: P(first attempt correct) + P(second attempt correct & first attempt wrong)
Which is: P(first attempt correct) + P(first attempt wrong) P(second attempt correct | first attempt wrong)  by definition of conditional probability.

And also what is the sample space and what are the correct events?

It depends on how you are guessing.  Let us say, there are $n$ possible (but only one correct), and you are selecting up to two without bias or repetition (basically: drawing without replacement).  Then the probability of obtaining the correct answer is:
$$\tfrac {1}{~n~}+\tfrac{~n-1~}{n}\cdot\tfrac{1}{~n-1~} ~=~ \tfrac {2}{~n~}$$
