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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

Total number of object= 100+50=150 n(S)=^(150) C_2 =11175 Total number of rusted= (50%×100)+(50%×50)=75 E_1 =event of select 2 bolt out 100 n(E_1 )=^(100) C_2 =4950 P(E_1 )=((4950)/(11175))=0.4430 E_2 =event of select 2 rusted out 75 n(E_2 )=^(75) C_2 =2775 P(E_2 )=((2775)/(11175))=0.2483 P(E_1 ∩E_2 )=((n(E_1 ∩E_2 ))/(n(S)))=((^(50) C_2 )/(^(150) C_2 )) P(E_1 or E_2 )=P(E_1 ∪ E_2 )= P(E_1 )+P(E_2 )−P(E_1 ∩ E_2 ) ((4950+2775−1225)/(11175))=0.58

$$\mathrm{Total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{object}= \\ $$$$\mathrm{100}+\mathrm{50}=\mathrm{150} \\ $$$$\mathrm{n}\left(\mathrm{S}\right)=^{\mathrm{150}} \mathrm{C}_{\mathrm{2}} \:=\mathrm{11175} \\ $$$$\mathrm{Total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{rusted}= \\ $$$$\left(\mathrm{50\%}×\mathrm{100}\right)+\left(\mathrm{50\%}×\mathrm{50}\right)=\mathrm{75} \\ $$$$\mathrm{E}_{\mathrm{1}} =\mathrm{event}\:\mathrm{of}\:\mathrm{select}\:\mathrm{2}\:\mathrm{bolt}\:\mathrm{out}\:\mathrm{100} \\ $$$$\mathrm{n}\left(\mathrm{E}_{\mathrm{1}} \right)=^{\mathrm{100}} \mathrm{C}_{\mathrm{2}} =\mathrm{4950} \\ $$$$\mathrm{P}\left(\mathrm{E}_{\mathrm{1}} \right)=\frac{\mathrm{4950}}{\mathrm{11175}}=\mathrm{0}.\mathrm{4430} \\ $$$$\mathrm{E}_{\mathrm{2}} =\mathrm{event}\:\mathrm{of}\:\mathrm{select}\:\mathrm{2}\:\mathrm{rusted}\:\mathrm{out}\:\mathrm{75} \\ $$$$\mathrm{n}\left(\mathrm{E}_{\mathrm{2}} \right)=^{\mathrm{75}} \mathrm{C}_{\mathrm{2}} =\mathrm{2775} \\ $$$$\mathrm{P}\left(\mathrm{E}_{\mathrm{2}} \right)=\frac{\mathrm{2775}}{\mathrm{11175}}=\mathrm{0}.\mathrm{2483} \\ $$$$\mathrm{P}\left(\mathrm{E}_{\mathrm{1}} \cap\mathrm{E}_{\mathrm{2}} \right)=\frac{\mathrm{n}\left(\mathrm{E}_{\mathrm{1}} \cap\mathrm{E}_{\mathrm{2}} \right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{\:^{\mathrm{50}} \mathrm{C}_{\mathrm{2}} }{\:^{\mathrm{150}} \mathrm{C}_{\mathrm{2}} \:} \\ $$$$\mathrm{P}\left(\mathrm{E}_{\mathrm{1}} \mathrm{or}\:\mathrm{E}_{\mathrm{2}} \right)=\mathrm{P}\left(\mathrm{E}_{\mathrm{1}} \cup\:\mathrm{E}_{\mathrm{2}} \right)= \\ $$$$\mathrm{P}\left(\mathrm{E}_{\mathrm{1}} \right)+\mathrm{P}\left(\mathrm{E}_{\mathrm{2}} \right)−\mathrm{P}\left(\mathrm{E}_{\mathrm{1}} \cap\:\mathrm{E}_{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{4950}+\mathrm{2775}−\mathrm{1225}}{\mathrm{11175}}=\mathrm{0}.\mathrm{58} \\ $$

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