Mel divided dozens of cookies between 7 classrooms, each
2 answers:
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Answer:
Step-by-step explanation:
Suppose we think of an alphabet X to be the Event of the evidence.
Also, if Y be the Event of cheating; &
Y' be the Event of not involved in cheating
From the given information:
Thus,
P(Y') = 1 - 0.01
P(Y') = 0.99
The probability of cheating & the evidence is present is = P(YX)
The probabilities of not involved in cheating & the evidence are present is:
(b)
The required probability that the evidence is present is:
P(YX or Y'X) = 0.006 + 0.000099
P(YX or Y'X) = 0.006099
(c)
The required probability that (S) cheat provided the evidence being present is:
Using Bayes Theorem