Respuesta :

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[tex]Pr(x)=\sum_{x\mathop{=}0}^nnC_x\times p^x\times q^{n-x}[/tex][tex]x=0,\text{ n=43, p=}\frac{1}{124},\text{ q=1-}\frac{1}{124}=\frac{123}{124}[/tex]

[tex]Pr(x=0)=43C_0\frac{}{}\times(\frac{1}{124})^0\times(\frac{123}{124})^{43}=0.71[/tex]

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