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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 E1= {S}, E2= {FS}, E3= {FFS} and so on .Suppose the probability of any particular battery being satisfactory is 0.99. All Ei are disjoints. Write down complete sample space and prove that P (sample space) =1.

User Jen Grant
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1 Answer

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It's clear enough that


\mathbb P(E_i)=\mathbb P(i=j)=\begin{cases}0.99&\text{for }j=1\\0.01*0.99&\text{for }j=2\\0.01^2*0.99&\text{for }j=3\\\vdots\end{cases}

Because each of the
E_i are disjoint, it follows that


\displaystyle\mathbb P(\text{sample space})=\mathbb P(E_1\cup E_2\cup\cdots)=\mathbb P\left(\bigcup_(i=1)^\infty E_i\right)=\sum_(i=1)^\infty\mathbb P(E_i)

=0.99+0.01*0.99+0.01^2*0.99+\cdots

=0.99(1+0.01+0.01^2+\cdots)

=0.99\displaystyle\sum_(n=0)^\infty 0.01^n

=0.99*\frac1{1-0.01}

=(0.99)/(0.99)=1
User Montoya
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