Ukucinga ukuba inani \(Y\) labo bonke abantu abakhe baba kwaye ekugqibeleni baya kuzalwa lilinganiselwe, makhe \(x\) ibe yindawo yakho epheleleyo ukususela ekuqaleni koluhlu. Emva koko \(0 < \frac{x}{Y} \leq 1\) . Ngoku sinokuthi ngokunokwenzeka ukuba \(95\%\) uphakathi kwabokugqibela \(95\%\) kubo bonke abantu abakhe bazalwa, ngoko \(0,05 < \frac{x}{Y} \leq 1\) kwaye ke ngoko \(Y < \frac{x}{0,05} = \frac{100 \cdot x}{5} = 20 \cdot x\) .
Ngokoqikelelo \(x \approx 6 \cdot 10^{10}\) kwaye ke ngoko \(Y < 120 \cdot 10^{10}\) . Ukuba ubude bobomi buhlala bufana kwaye nenani labantu abaphila ngexesha elifanayo lizinzile, kusekho malunga ne \(10.000\) iminyaka eseleyo kwi \(Yx = 114 \cdot 10^{10}\) . Ingxoxo yomhla wentshabalalo iyasebenza ngokulinganayo kuzo zonke iindawo kwimbali - umntu ungenza ingxoxo efanayo \(2000\) kwiminyaka eyadlulayo okanye \(5000\) kwiminyaka ezayo; ingqiqo esisiseko isasebenza (umda ophezulu we \(Y\) uba mkhulu ngokufanelekileyo).
Olu lingelo lwengcinga lulandelayo lusebenza ngendlela efanayo: Qwalasela iiurns ezimbini \(A\) ezine \(100\) iibhola kunye \(B\) kunye \(100\) neebhola ezigidi. Awuyazi ukuba yeyiphi i-urn. Ukuba ngoku uzobe ngokungaboniyo ukusuka kwenye yezi urns ezimbini kwaye ufumane ibhola ngenombolo \(42\) , kunokwenzeka ngakumbi ukuba uvela urn \(A\) kunokuba urn \(B\) (kusenokwenzeka kakhulu ukuba uya kuba phakathi kwabokugqibela \(95\%\) kubo bonke abantu abake bazalwa, kwaye akunakwenzeka kakhulu ukuba uya kuba phakathi kwabo bonke abantu bokuqala ukuzalwa \(5\%\) .
Ngoko ke i-urn ihlala igcwalisa iibhola ezintsha ekuhambeni kwexesha, kwaye ukukhupha inombolo nangaliphi na ixesha ngexesha lisixelela into malunga nenani elinokwenzeka leebhola ngelo xesha, kodwa akukho nto malunga nenani lebhola kwixesha elizayo. urn. Oku kuya kufuna uhlalutyo lwe-urn.