312-50V10 Question #486: Real Exam Question with Answer & Explanation

The correct answer is A: $62.5. ALE is calculated as SLE multiplied by ARO. With an asset value of $500, an EF of 0.5, and a failure rate of once every four years, the result is $62.50.

Question

The chance of a hard drive failure is known to be once every four years. The cost of a new hard drive is $500. EF (Exposure Factor) is about 0.5. Calculate for the Annualized Loss Expectancy (ALE).

Options

A$62.5
B$250
C$125
D$65.2

Explanation

ALE is calculated as SLE multiplied by ARO. With an asset value of $500, an EF of 0.5, and a failure rate of once every four years, the result is $62.50.

Common mistakes.

B. $250 represents only the SLE (asset value x EF) and omits dividing by the four-year recurrence period to annualize the figure.
C. $125 results from incorrectly using an ARO of 0.5 (once every two years) instead of the stated once every four years (0.25).
D. $65.2 does not correspond to any correct application of the ALE formula using the values provided.

Concept tested. Annualized Loss Expectancy (ALE) calculation

Reference. https://csrc.nist.gov/publications/detail/sp/800-30/rev-1/final

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