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\subsection*{4.3}
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Our random value is $R = u-bit long digit$ which means that it has $2^u$ possible values.
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And since we know that $q$ balls into $p$ holes a colision is bound to happend at the probability of $\frap{q^2}{2p}$ we can calculate:
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And since we know that $q$ balls into $p$ holes a colision is bound to happend at the probability of $\frac{q^2}{2p}$ we can calculate:
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$$\frac{q^2}{2(2^u)}\gt\frac{1}{2}\iff q\gt 2^{\frac{u}{2}}$$
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\subsection*{4.4}
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