SOLUTION: can someone help me with this, i dont understand the question suppose that the probability that a student passes a test the fisrt time is 0.6. for those who fail the first time

Question 1088308: can someone help me with this, i dont understand the question
suppose that the probability that a student passes a test the fisrt time is 0.6. for those who fail the first time, the probability of passing the test the second time is 0.8
a) determine the probability that randomly selected student passes the test
b) if student passes the test determine the probability that she or he did on the first time
Answer by natolino_2017(77) (Show Source): You can put this solution on YOUR website!
a) P(Random Student Passed the Test) = P(passed the test on the first attempt) + P(Passed the test on the second attempt).
The Second probability is not 0.8, because that's the Probability with the condition that you already failed the first attempt.
P(Random Student Passed the Test) = 0.6 + (1 - 0.6)(0.8) = 0.92 = 92%.
b) P(Passed the test on the first attempt/ Passed the test) = P(Passed the test/Passed the test on the first attempt)/(P(Passed the Test)) (Bayes's Formula)
= 0.6/(0.92) = 15/23 = 65.22%
Another way of seeing a) and b) is to imagine that the class has 100 Students, which 60 passed the test of the first attempt, then 40 take the second test and 32 passed.
So the result is that 8 students fail the class, and 92 passed (60 on the first and 32 on the second). So the a) is 92/100 =92%.
for b) the universe has changed, we no longer have 100 student, we only have 92 student, which they passed the test.
so b) is 60/92 = 65.22%
@natolino_

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