Answer:
0.7385 = 73.85% probability that it is indeed a sample of copied work.
Explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
![P(B|A) = (P(A \cap B))/(P(A))](https://img.qammunity.org/2022/formulas/mathematics/college/r4cfjc1pmnpwakr53eetfntfu2cgzen9tt.png)
In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Identified as a copy
Event B: Is a copy
Probability of being identified as a copy:
80% of 15%(copy)
100 - 95 = 5% of 100 - 15 = 85%(not a copy). So
![P(A) = 0.8*0.15 + 0.05*0.85 = 0.1625](https://img.qammunity.org/2022/formulas/mathematics/college/c9rf0f6iuz8qfr20vh00sddji4mk4ibr8x.png)
Probability of being identified as a copy and being a copy.
80% of 15%. So
![P(A \cap B) = 0.8*0.15 = 0.12](https://img.qammunity.org/2022/formulas/mathematics/college/ssw3f4hn2olfgcv1djpzto2v6gl95lauuc.png)
What is the probability that it is indeed a sample of copied work?
![P(B|A) = (P(A \cap B))/(P(A)) = (0.12)/(0.1625) = 0.7385](https://img.qammunity.org/2022/formulas/mathematics/college/7i6xif2ggy4s8s4csk8p21ikw9isjwbx4w.png)
0.7385 = 73.85% probability that it is indeed a sample of copied work.