Answer:
0.7385 = 73.85% probability that it is indeed a sample of copied work.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] 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
[tex]P(A) = 0.8*0.15 + 0.05*0.85 = 0.1625[/tex]
Probability of being identified as a copy and being a copy.
80% of 15%. So
[tex]P(A \cap B) = 0.8*0.15 = 0.12[/tex]
What is the probability that it is indeed a sample of copied work?
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.12}{0.1625} = 0.7385[/tex]
0.7385 = 73.85% probability that it is indeed a sample of copied work.