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Here m=6 and k=8

Probability of slot not hashed = $\frac{m-1}{m}$

Probability that particular slot being empty after k insertion =$\begin{pmatrix} \frac{m-1}{m} \end{pmatrix}^{k}$

So not being empty for particular slot=$1- \begin{pmatrix} \frac{m-1}{m} \end{pmatrix}^{k}$

Each slot having same probability so= $m \times \begin{pmatrix} 1- \begin{pmatrix} \frac{m-1}{m} \end{pmatrix}^{k} \end{pmatrix}$

So ans should be: $6 \times \begin{pmatrix} 1- \begin{pmatrix} \frac{5}{6} \end{pmatrix}^{8} \end{pmatrix}$ = 4.6045
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