Statistics for Maintenance Modelling: Basic Probability Concepts

# Example of Proposition 3¤

| Suppose that we toss two coins. The sample space is the set of all possible outcomes, i.e. S = {(H, H), (H, T), (T, H), (T, T)}. Let's assume that each of the four points in the sample space is equally likely and hence has probability of 1/4. If E is the event that the first coin falls heads and F is the event that the second coin falls heads, we can write:

E = {(H, H), (H, T)} and F = {(H, H), (T, H)}

From proposition 3 we have that the probability that either the first or the second coin falls heads is given by:

P(EF) = P(E) + P(F) - P(EF)

= P({(H, H), (H, T)}) + P({(H, H), (T, H)}) - P({(H, H)})

= (1/4 1/4) + (1/4 1/4) - 1/4

= 3/4

This probability could have been computed directly because:

P(EF) = P({(H, H), (H, T), (T, H)}) = 3/4

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