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dsp73
3 years ago
8

651,750 nearest ten thousand

Mathematics
1 answer:
Levart [38]3 years ago
7 0

Answer:

650,000

Step-by-step explanation:

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

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

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

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The required probability that the evidence is present is:

P(YX  or Y'X) = 0.006 + 0.000099

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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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Answer:


Step-by-step explanation:


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