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Openai/6843075e-8924-8006-8475-25672d1f30d4
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===== | | | ===== | --- | --- | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Multiply them all: P=11000Γ11000Γ1500Γ1500Γ11000Γ11000Γ1500Γ1300Γ1200Γ1500Γ1200Γ150Γ150P = \frac{1}{1000} \times \frac{1}{1000} \times \frac{1}{500} \times \frac{1}{500} \times \frac{1}{1000} \times \frac{1}{1000} \times \frac{1}{500} \times \frac{1}{300} \times \frac{1}{200} \times \frac{1}{500} \times \frac{1}{200} \times \frac{1}{50} \times \frac{1}{50}P=10001βΓ10001βΓ5001βΓ5001βΓ10001βΓ10001βΓ5001βΓ3001βΓ2001βΓ5001βΓ2001βΓ501βΓ501β Letβs calculate: P=110002β 5003β 10002β 300β 2002β 502P = \frac{1}{1000^2 \cdot 500^3 \cdot 1000^2 \cdot 300 \cdot 200^2 \cdot 50^2}P=10002β 5003β 10002β 300β 2002β 5021β P=1106β 1.25Γ108β 9Γ104β 2500P = \frac{1}{10^{6} \cdot 1.25 \times 10^8 \cdot 9 \times 10^4 \cdot 2500}P=106β 1.25Γ108β 9Γ104β 25001β Just a rough approximation gives us: Pβ11026=0.00000000000000000000000001P \approx \frac{1}{10^{26}} = 0.00000000000000000000000001Pβ10261β=0.00000000000000000000000001
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