EX 2.12 test batteries

Consider testing batteries coming off an assembly line one by one until one having a voltage within prescribed limits is found. The simple events are

• $E_1=\{S\}$,
• $E_2=\{FS\}$,
• $E_3=\{FFS\}$,
• $E_4=\{FFFS\}$, …

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Suppose the probability of any particular battery being satisfactory is .99. Then it can be shown that

• $P\left(E_1\right)=.99$
• $P\left(E_2\right)=\left(.01\right)\left(.99\right)$
• $P\left(E_3\right)={\left(.01\right)}^2\left(.99\right),\dots$ is an assignment of probabilities to the simple events that satisfies the axioms. In particular, because the $E_i$‘s are disjoint and $\mathcal{S}=E_1\cup E_2\cup E_3\cup \dots$ , it must be the case that

$$\begin{array}{rl}1& =P\left(\mathcal{S}\right)\\ & =P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)+\cdots \\ & =.99\left[1+.01+{\left(.01\right)}^2+{\left(.01\right)}^3+\cdots \right]\end{array}$$

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Here we have used the formula for the sum of a geometric series:

$$a+ar+a{r}^2+a{r}^3+\cdots =\frac{a}{1-r}$$

However, another legitimate (according to the axioms) probability assignment of the same “geometric” type is obtained by replacing. 99 by any other number $p$ between 0 and 1 (and .01 by $1-p$ ).

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