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.

Answer 1

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

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

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

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