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 1 I-Worm.Klez 46.30% 2 I-Worm.Lentin 17.83% 3 I-Worm.Tanatos 5.21% 4 I-Worm.BadtransII 1.30% 5 Macro.Word97.Thus 1.80% 6 I-Worm.Hybris 0.11% 7 I-Worm.Bridex 0.35% 8 I-Worm.Magistr 0.02% 9 Win95.CIH 0.24% 10 I-Worm.Sircam 26.84% Suppose that 20 malicious software instances are reported. Assume that the malicious sources can be assumed to be independent. (a) What is the probability that at least one instance is "Klez"? Round your answer to four decimal places (e.g. 98.7654). (b) What is the probability that five or more instances are "Klez"? Round your answer to four decimal places (e.g. 98.7654). (c) What is the mean of the number of "Klez" instances among the 20 reported? Round your answer to two decimal places (e.g. 98.76). (d) What is the standard deviation of the number of "Klez" instances among the 20 reported? Round your answer to two decimal places (e.g. 98.76).

Answer 1

Answer:

a) 99.9999%

b) 98.5938%

c) 9.26

d) 2.23

Step-by-step explanation:

A="at least one instance is Klez"

B="Five or more instances are Klez"

Xi={1 if The i-th instance is Klez, 0 else}

Y=Bin(20, 0.463)

a) $P(A)=P(Y\geq 1 )=1-P(Y=0)=1-P(\displaystyle\bigcap_{i=1}^{20}{X_i=1})=1- P(\displaystyle\sum_{i=1}^{20}X_i=20)=1- \displaystyle\prod_{i=1}^{20} P(X_i=1)=1-P(X_i=1)^{20}=1-0.4630^{20}=0.9999$

b) $P(B)=P(Y\geq 5)=1-P(Y<5)=1-\displaystyle\sum_{n=0}^{4}P(Y=n)=1- P(\displaystyle\sum_{i=1}^{20}X_i<5)$

$P(B)=1-\displaystyle\sum_{n=0}^{4}(^{20}_n)(0.4630)^n(1-0.4630)^{20-n} = 1-(3.9763\times10^{-6}+6.8568\times10^{-5}+5.6163\times10^{-4}+2.9054\times10^{-3}+0.0106)=0.9859$

c) $E[Y]=20\times0.463=9.26$

d) $Var(Y)=20\times0.463\times(1-0.463)=4.97, \sigma=\sqrt{4.97}=2.23$