# Example 37 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 15, 2020 by Teachoo

Last updated at Feb. 15, 2020 by Teachoo

Transcript

Example 37 If a machine is correctly set up, it produces 90% acceptable items. If it is incorrectly set up, it produces only 40% acceptable items. Past experience shows that 80% of the set ups are correctly done. If after a certain set up, the machine produces 2 acceptable items, find the probability that the machine is correctly setup.Let E1 : Event that the machine is correctly setup E2 : Event that the machine is incorrectly setup A : Event that the Machine produce two acceptable items We need to find out the probability that the machine have a correct set up if it produce two acceptable item i.e. P(E1|A) P(E1|A) = (๐(๐ธ_1 ).๐(๐ด|๐ธ_1))/(๐(๐ธ_1 ).๐(๐ด|๐ธ_1)+๐(๐ธ_2 ).๐(๐ด|๐ธ_2) ) "P(E1)" = Probability that machine is correctly setup = 80% = 80/100 =0.8 P(A|E1) = Probability that machine produce 2 acceptable items if it have correct set up = ๐๐% ร ๐๐% = 90/100 ร 90/100 = 0.9 ร 0.9 = 0.81 "P(E2)" = Probability that machine is incorrectly setup = (100โ80)% = 20%=20/100 = 0.2 P(A|E1) = Probability that machine produce 2 acceptable items if it have incorrect set up = ๐๐% ร ๐๐% = 40/100ร40/100 = 0.4 ร 0.4 = 0.16 Putting values in the formula P(E1|A) = (๐(๐ธ_1 ).๐(๐ด|๐ธ_1))/(๐(๐ธ_1 ).๐(๐ด|๐ธ_1)+๐(๐ธ_2 ).๐(๐ด|๐ธ_2) ) = (0.8 ร ๐.๐๐)/(0.8 ร ๐.๐๐ + 0.2 ร ๐.๐๐) = 0.648/0.680 = 0.95

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Example 28 Important Deleted for CBSE Board 2022 Exams

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Chapter 13 Class 12 Probability (Term 2)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.