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Question Number 168179 by peter frank last updated on 05/Apr/22

Answered by peter frank last updated on 07/Apr/22

$$\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{object}=$$$$100+50=150$$$$\mathrm{n}\left(\mathrm{S}\right){=}^{150}{\mathrm{C}}_2=11175$$$$\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{rusted}=$$$$\left(50\mathrm{\%}\times 100\right)+\left(50\mathrm{\%}\times 50\right)=75$$$${\mathrm{E}}_1=\mathrm{event}\mathrm{of}\mathrm{select}2\mathrm{bolt}\mathrm{out}100$$$$\mathrm{n}\left({\mathrm{E}}_1\right){=}^{100}{\mathrm{C}}_2=4950$$$$\mathrm{P}\left({\mathrm{E}}_1\right)=\frac{4950}{11175}=0.4430$$$${\mathrm{E}}_2=\mathrm{event}\mathrm{of}\mathrm{select}2\mathrm{rusted}\mathrm{out}75$$$$\mathrm{n}\left({\mathrm{E}}_2\right){=}^{75}{\mathrm{C}}_2=2775$$$$\mathrm{P}\left({\mathrm{E}}_2\right)=\frac{2775}{11175}=0.2483$$$$\mathrm{P}({\mathrm{E}}_1\cap {\mathrm{E}}_2)=\frac{\mathrm{n}({\mathrm{E}}_1\cap {\mathrm{E}}_2)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{{}^{50}{\mathrm{C}}_2}{{}^{150}{\mathrm{C}}_2}$$$$\mathrm{P}\left({\mathrm{E}}_1\mathrm{or}{\mathrm{E}}_2\right)=\mathrm{P}({\mathrm{E}}_1\cup {\mathrm{E}}_2)=$$$$\mathrm{P}\left({\mathrm{E}}_1\right)+\mathrm{P}\left({\mathrm{E}}_2\right)-\mathrm{P}({\mathrm{E}}_1\cap {\mathrm{E}}_2)$$$$\frac{4950+2775-1225}{11175}=0.58$$

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