Exam question

Question

The probability of the student solving Q1 is 60% . If he solves it the probability of him passing the exam is 80% , if he doesn’t then the probability of failing is 55%.After the exam the student has passed,what is the probability that he solved Q1 correctly?

 

 

i had this question on my last exam . I tried solving it using the conditional rule then after getting the probability of passing i used bayes rule to get the probability of solving Q1 given the student has passed. i got 0.71 .However i failed this subject and i got 0 points from this question ,can any body solve it please?

Answer 1

0.6------------------Q1 Correct------------------------0.80 Of Passed Test

0.4------------------Q1 Wrong-------------------------0.55 Of Failed Test

0.6------------------Q1 Correct------------------------0.20 Of Failed Test

0.4------------------Q1 Wrong-------------------------0.45 Of Passed Test

The probability of solving Q1 or P(Q1 Correct) is given as 0.6.

The probability of missing Q1 or P(Q1 Wrong) is then equal to 0.4.

The probability of passing the test given solution of Q1 or P(Test Passed|Q1 Correct) is given as 0.8.

The probability of failing the test given solution of Q1 or P(Test Failed|Q1 Correct) is then 0.2.

The probability of failing the test given no solution of Q1 or P(Test Failed|Q1 Wrong) is given as 0.55.

The probability of passing the test given no solution of Q1 or P(Test Passed|Q1 Wrong) is then 0.45.

Then write: P(Q1 Correct|Test Passed) is equal to

P(Q1 Correct)P(Test Passed|Q1 Correct) Divided By

[P(Q1 Correct)P(Test Passed|Q1 Correct) + P(Q1 Wrong)P(Test Failed|Q1 Wrong)+P(Q1 Correct)P(Test Failed|Q1 Correct)+P(Q1 Wrong)P(Test Passed|Q1 Wrong)].

Place values to obtain:

P(Q1 Correct|Test Passed) = (0.6)(0.8) Divided By [(0.6)(0.8) + (0.4)(0.55) + (0.6)(0.2) + (0.4)(0.45)]

which reduces to 0.48/1 or 0.48.