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Therefore, the marginal probability density function of Y1 is fY1 1 y1 2 fX1 1 y1 y2 2 fX2 1 y2 2 dy2

MD5 attack, we give an example that shows how a meaningless collision can be used to break security in a very meaningful way.

The notation is simpler if the variable of integration y2 is replaced with x and y1 is replaced with y. Then the following result is obtained.

P(E) = 30(0.01) = 0.30

I can t explain myself, I m afraid, Sir, said Alice,

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It is frequently necessary to assign probabilities to events that are composed of several outcomes from the sample space. This is straightforward for a discrete sample space. EXAMPLE 2-9 Assume that 30% of the laser diodes in a batch of 100 meet the minimum power requirements of a speci c customer. If a laser diode is selected randomly, that is, each laser diode is equally likely to be selected, our intuitive feeling is that the probability of meeting the customer s requirements is 0.30. Let E denote the subset of 30 diodes that meet the customer s requirements. Because E contains 30 outcomes and each outcome has probability 0.01, we conclude that the probability of E is 0.3. The conclusion matches our intuition. Figure 2-10 illustrates this example. For a discrete sample space, the probability of an event can be de ned by the reasoning used in the example above.

De nition For a discrete sample space, the probability of an event E, denoted as P(E), equals the sum of the probabilities of the outcomes in E.

EXAMPLE 2-10

Message Digest 5, or MD5, is a strengthened version of MD4. Like MD4, the MD5 hash was invented by Rivest [122]. Also, MD5 was obviously used as the model for SHA-1, since they share many common features. It is undoubtedly the case that MD5 and SHA-1 are the two most widely used hash algorithms today, but use of MD5 will certainly decline over time, since it is now considered broken. At the time of this writing, it appears certain that SHA-1 will soon share the same fate as MD5.

A random experiment can result in one of the outcomes {a, b, c, d} with probabilities 0.1, 0.3, 0.5, and 0.1, respectively. Let A denote the event {a, b}, B the event {b, c, d}, and C the event {d}.Then, P1A2 P1B2 P1C2 0.1 0.3 0.1 0.3 0.5 0.4 0.1 0.9

Also, P1A 2 0.6, P1B 2 0.1, and P1C 2 0.9. Furthermore, because A B 5b6, P1A B2 0.3. Because A B 5a, b, c, d6, P1A B2 0.1 0.3 0.5 0.1 1. Because A C is the null set, P1A C2 0 .

MD5 is similar to MD4 and in this section, we often refer to our previous discussion of MD4. It is important to read Section 5.3 carefully before attempting this section, since both the operation of MD4 and various aspects of the MD4 attack will be referenced here.

EXAMPLE 2-11

A visual inspection of a location on wafers from a semiconductor manufacturing process resulted in the following table:

52 52 53 53 53 54 54 54 54 55 55 56 56 57 57 58 59 60 61 61 61 61 63

If one wafer is selected randomly from this process and the location is inspected, what is the probability that it contains no particles If information were available for each wafer, we could de ne the sample space as the set of all wafers inspected and proceed as in the example with diodes. However, this level of detail is not needed in this case. We can consider the sample space to consist of the six categories that summarize the number of contamination particles on a wafer. Then, the event that there is no particle in the inspected location on the wafer, denoted as E, can be considered to be comprised of the single outcome, namely, E {0}. Therefore, P1E2 0.4

F ( A ,B, ) = ( AA B ) V ( 1 AA C ) C G ( A ,B, ) = ( AA C ) V ( B A -C) C H(A,B,C) A @ B @ C = I ( A ,B, C) = B @ ( AV -C)

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