How to find probability from given sample space ?

Question

In answering on a multiple choice test, a student either know the answer or guesses. Let p be the probability that the students knows the answer and 1-p be the probability that the student guesses. Assume that a student who guesses at the answer will be correct with probability 1/m, where m is the number of multiple choice alternatives. what is the probability that he answers the question correctly ?

In this ques sample space will be { KC ,GC ,GI} 

K= knowing answer , C = answer is correct , G= guesses the answer , I=Incorrect answer 

My confusion is that in the question P(K)=p  but shouldn't it be 1/3 since we have  2 mutually exclusive events here 

1. Knowing and answers correctly in one set, so P(K)=1/3 

2. Guessing the answer which may be correct or incorrect , so P(G)=2/3 

why is it p in the question ?

Comments

I think  "p be the probability that the students knows the answer " states that out of 10 question he knows 2 .

Answer : P(k) + P(GC) = p+(1-p)* 1/m

Answer 1

"If the student guesses then he does not know the answer & if the student knows the answer then he is not going to guess."

So yes $P(K)$ and $P(G)$ are mutually exclusive events, since only one of them can happen at a time.

But given $P(K)$ and $P(G)$ are mutually exclusive events, the only fact you can guarantee here is 

$P(K) \bigcap P(G) = 0$ and nothing more.

You can not infer the exact values of $P(K)$ and $P(G)$ from their mutual exclusiveness.

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It would be something like:

you have two sets $A$ & $B$, and it is given that 

$A \cup\ B = 1$ and $A \cap\ B = 0$.

What can you say about sets $A$ & $B$ ?

From $(A \cup B)= A + B + (A \cap B)$

we can say that, $ A + B = 1$

Clearly we can not conclude from this equation that what will be the individual values of $A$ & $B$.

But one thing about which you are sure is if you know the cardinality of $A$(say the cardinality of $A$ is $x$), then you can find the cardinality of $B$(it will be $( 1 - x)$), and vice versa.

Similarly here with the given information about $P(K)$ and $P(G)$ you can only say that 

$P(K) + P(G) = 1$

Which shows that events $K$ and $G$ are totally exhaustive, that is the student MUST attempt the question no matter whether his  answers is correct or incorrect but he can not leave it unattempted.

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Coming to the answer of the original question, Umang is right.

In your conventions,

$P(KC) = p$,

$P(GC) = (1 - p) \cdot \frac{1}{m}$, since out of $m$ choices only one is correct.

$P(GI) = (1 - p) \cdot \frac{(m - 1)}{m}$, since $(m - 1)$ are incorrect out of $m$ choices.

$P(C) = \frac {P(KC) + P(GC)}{P(KC) + P(GC) + P(GI)}$

but $P(KC) + P(GC) + P(GI) = 1$

Hence the probability that he answers the question correct would be

$P(C) = P(KC) + P(GC) = p + (1 - p) \cdot \frac{1}{m}$