Question

Please please help me out

Answer 1

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`Pr(x)=\sum_{x\mathop{=}0}^nnC_x* p^x* q^(n-x)`
`x=0,\text{ n=43, p=}(1)/(124),\text{ q=1-}(1)/(124)=(123)/(124)`

`Pr(x=0)=43C_0\frac{}{}*((1)/(124))^0*((123)/(124))^(43)=0.71`