Respuesta :

LammettHash
[tex]\mathbb P(9\le X\le 10)=\mathbb P(X=9)+\mathbb P(X=10)[/tex]

Since

[tex]\mathbb P(X=x;n,p)=\dbinom nxp^x(1-p)^{n-x}[/tex]

you have

[tex]\mathbb P(9\le X\le10)=\dbinom{20}90.4^9(1-0.4)^{20-9}+\dbinom{20}{10}0.4^{10}(1-0.4)^{20-10}[/tex]
[tex]\mathbb P(9\le X\le10)=\dbinom{20}90.4^90.6^{11}+\dbinom{20}{10}0.4^{10}0.6^{10}[/tex]
[tex]\mathbb P(9\le X\le10)\approx0.2769[/tex]

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