Question

An article in Information Security Technical Report ["Malicious Software-Past, Present and Future" (2004, Vol. 9, pp. 6-18)] provided the following data on the top ten malicious software instances for 2002. The clear leader in the number of registered incidences for the year 2002 was the Internet worm "Klez," and it is still one of the most widespread threats. This virus was first detected on 26 October 2001, and it has held the top spot among malicious software for the longest period in the history of virology Place Name % Instances 51.45% 15.56% 1 I-Worm.Klez 2 I-Worm.Lentin 3 1-Worm.Tanatos 5.20% 4 I-worm.Badtransli 1.97% 5 Macro.word97.Thus 1.63% 6 7 I-Worm.Hybris I-Worm.Bridex 0.19% 0.39% 8 I-worm .Magistr 0.34% 9 Win95.CIH 10 I-Worm.Sircam 0.48% 22.79% Suppose that 20 malicious software instances are reported. Assume that the malicious sources can be assumed to be independent

Answer 1

(a)

The probability of instance of Klez is 0.5145

Let X be the number of instances of Klez in 20 malicious software instances. Then X ~ Binomial(n = 20, p = 0.5145)

Using binomial distribution, the probability that at least one instance is Klez = P(X $\ge$ 1)

= 1 - P(X < 1) = 1 - P(X = 0)

= 1 - ²⁰C₀ * 0.5145⁰ * (1 - 0.5145)^20-0

= 1 - 0.4855²⁰

= 0.9999995

$\approx$1 (Rounding to 4 decimal places)

(b)

Using binomial distribution, the probability that five or more instances is Klez = P(X $\ge$ 5)

= 1 - P(X < 5) = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)]

= 1 - [²⁰C₀ * 0.5145⁰ * (1 - 0.5145)^20-0 + ²⁰C₁ * 0.5145¹ * (1 - 0.5145)^20-1 + ²⁰C₂ * 0.5145² * (1 - 0.5145)^20-2 + ²⁰C₃ * 0.5145³ * (1 - 0.5145)^20-3 + ²⁰C₄ * 0.5145⁴ * (1 - 0.5145)^20-4 ]

= 1 - [0.4855²⁰ + 20 * 0.5145 *  0.4855¹⁹ + 190 * 0.5145² *  0.4855¹⁸ + 1140 * 0.5145³ *  0.4855¹⁷ + 4845 * 0.5145⁴ *  0.4855¹⁶ ]

= 1 - (0.004077882)

= 0.9959

(c)

Mean of number of "Klez" instances = np = 20 * 0.5145 = 10.29

(d)

Standard deviation of number of "Klez" instances =  $\sqrt{np(1-p)}$ = $\sqrt{20 * 0.5145 * (1-0.5145)}$

= 2.235128

$\approx$2.24 (Rounded to 2 decimal places)