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ikadub [295]
3 years ago
12

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.
Mathematics
1 answer:
sleet_krkn [62]3 years ago
5 0
It's clear enough that

\mathbb P(E_i)=\mathbb P(i=j)=\begin{cases}0.99&\text{for }j=1\\0.01\times0.99&\text{for }j=2\\0.01^2\times0.99&\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\times0.99+0.01^2\times0.99+\cdots
=0.99(1+0.01+0.01^2+\cdots)
=0.99\displaystyle\sum_{n=0}^\infty 0.01^n
=0.99\times\dfrac1{1-0.01}
=\dfrac{0.99}{0.99}=1
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