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Eight percent of all college graduates hired by companies stay with the same company for more than five years. The probability, rounded to four decimal places, that in a random sample of 14 such college graduates hired recently by companies, exactly 2 will stay with the same company for more than five years is _?_.

User Epieters
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P(X=2) &= {14\choose 2}(0.08)^2(0.92)^{12} \

&= \frac{14!}{2!(14-2)!}(0.08)^2(0.92)^{12} \


\sf\implies\:&=(14*13)/(2*1)(0.08)^2(0.92)^(12)

&= 91(0.08)^2(0.92)^{12} \

&\approx \boxed{0.2166


\begin{align}\huge\colorbox{black}{\textcolor{yellow}{\boxed{\sf{I\: hope\: this\: helps !}}}}\end{align}


\begin{align}\colorbox{black}{\textcolor{white}{\underline{\underline{\sf{Please\: mark\: as\: brillinest !}}}}}\end{align}


\textcolor{lime}{\small\textit{If you have any further questions, feel free to ask!}}


\huge{\bigstar{\underline{\boxed{\sf{\color{red}{Sumit\:Roy}}}}}}\\

User Tsvallender
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