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 vavanyo lwengcinga lulandelayo lusebenza ngendlela efanayo: Qwalasela ii-urns ezimbini \(A\) ezine \(100\) iibhola kunye \(B\) kunye \(100\) neebhola ezigidi. Awuyazi ukuba yeyiphi i-urn. Ukuba ngoku uzobe ngokungaboniyo ukusuka kwenye yee urn ezimbini kwaye ufumane ibhola enenombolo \(42\) , kunokwenzeka ukuba iphume kwi urn \(A\) kunokusuka kwi urn \(B\) (ikwanjalo kakhulu. kusenokwenzeka ukuba uphakathi kwabokugqibela \(95\%\) kubo bonke abantu abakhe bazalwa kwaye akunakwenzeka ukuba ube phakathi kwabokuqala \(5\%\) kubo bonke abantu abakhe bazalwa).
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.