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The probability that an employee will be late to work at a large corporation is 0.21. What is the probability on a given day that in a department of 5 employees 1 or 2 employees are late ?

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Let the probability that an eployee will be late to work be given by:


P(1|1)=0.21

Then, in a department of 5 employes, the problability that 1 or 2 employes are late is given by:


\begin{gathered} P((1\lor2)|5)=P(1|5)+P(2|5) \\ P(1|5)=P(1|1)\cdot(1-P(1|1))^4\cdot(5!)/(1!\cdot4!)=0.21\cdot0.79^4\cdot5=0.41 \\ P(2|5)=P(1|1)^2\cdot)(1-P(1|1))^3\cdot(5!)/(2!\cdot3!)=0.21^2\cdot0.79^3\cdot10=0.22 \\ \therefore P((1\lor2)|5)=0.41+0.22=0.63 \end{gathered}

User Michel Borges
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