A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheating student, e

Question

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheating student, evidence E is present with 60% percent probability, and in the case of a student that does not cheat, evidence E is present with a 0.01 percent probability. Suppose also that the proportion of students that cheat is 1 percent. Show all the steps including identification of what formulas/properties you used. Points will be deducted from answers if only the final answer is provided.

Required:
a. Determine the events, given probabilities and inferred probabilities.
b. Determine the probability that the evidence is present.
c. Determine the probability that S cheated given the evidence is present.

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Dâu 4 years 2021-08-13T04:20:41+00:00 1 Answers 32 views 0

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    2021-08-13T04:22:24+00:00

    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:

    P(\dfrac{X}{Y}) = 60\% = 0.6

    P(\dfrac{X}{Y'}) = 0.01\% = 0.0001

    P(Y) = 0.01

    Thus, P(Y') \ will\ be = 1 - P(Y)

    P(Y’) = 1 – 0.01

    P(Y’) = 0.99

    The probability of cheating & the evidence is present is = P(YX)

    P(YX) = P(\dfrac{X}{Y}) \ P(Y)

    P(YX) =0.6 \times 0.01

    P(YX) =0.006

    The probabilities of not involved in cheating & the evidence are present is:

    P(Y'X) = P(Y')  \times P(\dfrac{X}{Y'})

    P(Y'X) = 0.99  \times 0.0001 \\ \\  P(Y'X) = 0.000099

    (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

    P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}

    P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}

    P(\dfrac{Y}{X}) = 0.9838

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