Question

A student takes a multiple-choice exam. Suppose for each question he either knows the answer or guesses and chooses an option at random. Further suppose that if he knows the answer, the probability of a correct answer is 1, but if he guesses, the probability of a correct answer is 1/4. To pass, students need to answer at least 60% of the questions correctly. The student has "studied for a minimal pass," i.e., the student knows the answer to a question with probability 0.6. Given that he answers a correctly, what is the probability that he actually knows the answer? Group of answer choices

Answer 1

Answer:

The probability that he knows the answer given his answer is correct is 0.8571

Step-by-step explanation:

The problem can be solved by Bayes theorem

Let us take the Following Events

1) P(A)= probability that the student's answer is correct

2) P(B) = probability that the student knows the correct answer

For 2 events A and B Bayes theorem states that

The probability that an event 'B' occurs provided that event A has already occurred is given by "P(B|A)" as

$P(B|A)=\frac{P(A|B)\cdot P(B)}{P(A)}$

where

P(A|B) is the probability of event A to occur provided event B has occurred

from the given data we have

P(B) = 0.6

P(A) (probability that his answer is correct = probability that his answer is correct provided he knows the answer + Probability his answer is correct provided he gambles)

Thus

$P(A)=1\times 0.6+\frac{1}{4}\times 0.4=0.7$

Thus

$P(B|A)=\frac{1\times 0.6}{0.7}=0.8571$