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#1 2008-11-11 20:17:14

chetah
Member
Registered: 2008-02-15
Posts: 32

Probability

I am looking a probability question, the answer is given, but I cannot figure how they arrive at those figures.

An analysis of common household computers in a region reports that 33% of the computers of the computers are vulnerable to security penetrations through their telnet ports, 46% ftp ports, 73% printer ports.  18% are vulnerable through telnet and ftp ports, 36% through ftp and printer ports and 24% through telnet and printer ports and 12% through all their port.

Let's say their are 3 sets A, B and C. Let's say set A is telnet and B is Ftp and C printer.  The answer given is (for A only)

p(A) = 0.03
p(A) π p( B) = 0.06
p(A) π p(C) = 0.12
p(A) π p(B) π(C) = 0.12

what i found out is if i add up all the values above, it gives .33 (33/100).  My question how did they get from .33 to .03, .18 to .06 etc.

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#2 2008-11-13 17:46:37

Macy
Guest

Re: Probability

1. Think as you have 33 computers affected by virus in total
2. The numbers of computers have affected by 3 ports ( T, FTP, P) = 12
3. The number of computers have affected by 2 ports (T, P) = 12
4. The total number of computes affected by 2 ports (T , FTP) = 6
5. Total numbers of computers affected by port (T only) = 33- (numbers affected by 3 ports(T,FTP,P) + number affected by 2 ports (T,FTP) + affected by 2 ports (T,P) = 33 - (12 + 12+ 6) = 3

I suggest you draw a Venn diagram with 3 circle then you will get it

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