javeddehwar234
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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.

Respuesta :

LammettHash
It's clear enough that

[tex]\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}[/tex]

Because each of the [tex]E_i[/tex] are disjoint, it follows that

[tex]\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)[/tex]
[tex]=0.99+0.01\times0.99+0.01^2\times0.99+\cdots[/tex]
[tex]=0.99(1+0.01+0.01^2+\cdots)[/tex]
[tex]=0.99\displaystyle\sum_{n=0}^\infty 0.01^n[/tex]
[tex]=0.99\times\dfrac1{1-0.01}[/tex]
[tex]=\dfrac{0.99}{0.99}=1[/tex]

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