Question

第 5 个问题 a multiple choice exam has 10 questions. each question has 3 possible answers, of which one is correct. a student knows the correct answers to 4 questions and guesses the answers to the other 6 questions. it turns out that the student answered the first question correctly. what are the chances that the student was merely guessing?

Answer 1

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

$P(A/B)=\frac{P(A)*P(B/A)}{P(B)}$

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

$P(A)=\frac{4}{10} \\\\ P(B)=1-\frac{4}{10} =\frac{6}{10}$

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

$P(E/A)=1$

Probability that the answer is correct ,given that he guesses it is

$P(E/B)=\frac{1}{3}$   [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 $P(B/E)=\frac{P(B)*P(E/B)}{P(A)*P(E/A)+P(B)*P(E/B)}$

Substituting the values we get

$P(B/E)=\frac{\frac{6}{10} *\frac{1}{3} }{\frac{4}{10} *1+\frac{6}{10} *\frac{1}{3} }$

$=\frac{6}{30}*\frac{30}{18} \\ \\ =\frac{6}{18} \\ \\ =\frac{1}{3}$

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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