The chances that the student was merely guessing is 1/3.
Bayes Theorem determines the conditional probability of an event A given that event B has already occurred.
denoted by

let A be the event that the student knows the answer .
B be the event that the student does not knows the answer .
and
E be the event he gets answer correct .
According to the given question

Probability that the answer is correct ,given that he knows the answer is

Probability that the answer is correct ,given that he guesses it is
[as the MCQ has 3 options and only one is correct]
We need to find the probability that he guesses the answer given that it is correct.
Required probability 
Substituting the values we get


Therefore , the chances that the student was merely guessing is 1/3.
Learn more about Probability here brainly.com/question/13140147
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