I have made the following list out from the text below:
- 1% of the students who made their homework forgot to turn it in (A1)
- 99% of the students who made their homework remembered to turn it in. (B)
- 90% of the students completed their homework on time. (A2)
- 10% did not complete their homework on time. (A3)
- 5% of those who did not complete their homework on time lied and said they forgot to turn it in. (A4)
From here I am stuck on how to calculate it. I have seen the Bayes Theorem but I seem to have too many variables?
John tells his professor that he forgot to submit his homework assignment. From experience, the professor knows that students who finish their homework on time, actually forget to turn it in 1 in 100 times. She also knows that half of the students who have not finished their homework will falsely tell her they forgot to turn it in. She thinks that 90% of the students in this class completed their homework on time. What is the probability that Jack is telling the truth, i.e., he finished it but forgot to submit it?
Possible answers to chose from:
- a. 0.059
- b. 0.1525
- c. 0.556
Setup:
Let $A$ represent the event "The homework was actually completed by the student" and event $B$ the event "The homework was not turned in but the student says they completed it"
We are told the following: $Pr(A)=0.9$, $Pr(B\mid A)=0.01$, $Pr(B\mid A^c)=0.5$ which immediately then implies the additional $Pr(A^c)=0.1$, $Pr(B^c\mid A)=0.99$, $Pr(B^c\mid A^c)=0.5$
(note, $Pr(B\mid A^c)$ is $0.5$, not $0.05$ as per the phrase "She also knows that half of the students who have not finished their homework will falsely tell her they forgot to turn it in")
We wish to calculate $Pr(A\mid B)=\dfrac{Pr(B\mid A)Pr(A)}{Pr(B)}$
The next set of leading questions:
What piece or pieces of information from $Pr(A\mid B)=\dfrac{Pr(B\mid A)Pr(A)}{Pr(B)}$ are we still missing?
How might we go about calculating it?