For the sake of simplicity, first look at the case when you are interested in a unit that can only succeed or fail. For example, consider the case of a wine glass designed to withstand a fall of three feet onto a level cement surface.
The success/failure result of such a drop is determined by whether or not the glass breaks. [Note]
Furthermore, assume that:
Now given the above assumptions, the question of how the glass could be in the unreliable state just before trial can be answered in two mutually exclusive ways:
The first possibility is the probability of a successful trial, , where is the probability of failure in trial , while being in the unreliable state, , before the trial:
Secondly, the glass could have failed the trial, with probability , when in the unreliable state, , and having failed the trial, an unsuccessful attempt was made to fix, with probability
Therefore, the sum of these two probabilities, or possible events, gives the probability of being unreliable just before trial :
By induction, since :
To determine the probability of being in the reliable state just before trial , Eqn. (3) is subtracted from 1, therefore:
Define the reliability of the glass as the probability of not failing at trial . The probability of not failing at trial is the sum of being reliable just before trial , , and being unreliable just before trial but not failing , thus:
Now instead of , assume that the glass has some initial reliability or that the probability that the glass is in the unreliable state at , , then:
When , the reliability at the trial is larger than when it was certain that the device was unreliable at trial . A trend of reliability growth is observed in Eqn. (6). Let and , then Eqn. (6) becomes:
Eqn. (7) is now a model that can be utilized to obtain the reliability (or probability that the glass will not break) after the trial. Additional models, their applications and methods of estimating their parameters are presented in the following chapters.