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Sagot :
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
[tex]P(A/B)=\frac{P(A)*P(B/A)}{P(B)}[/tex]
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
[tex]P(A)=\frac{4}{10} \\\\ P(B)=1-\frac{4}{10} =\frac{6}{10}[/tex]
Probability that the answer is correct ,given that he knows the answer is
[tex]P(E/A)=1[/tex]
Probability that the answer is correct ,given that he guesses it is
[tex]P(E/B)=\frac{1}{3}[/tex] [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 [tex]P(B/E)=\frac{P(B)*P(E/B)}{P(A)*P(E/A)+P(B)*P(E/B)}[/tex]
Substituting the values we get
[tex]P(B/E)=\frac{\frac{6}{10} *\frac{1}{3} }{\frac{4}{10} *1+\frac{6}{10} *\frac{1}{3} }[/tex]
[tex]=\frac{6}{30}*\frac{30}{18} \\ \\ =\frac{6}{18} \\ \\ =\frac{1}{3}[/tex]
Therefore , the chances that the student was merely guessing is 1/3.
Learn more about Probability here https://brainly.com/question/13140147
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