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  1. Chapter 13 Class 12 Probability
  2. Serial order wise

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‘ƒ(๐ธ_1 ).๐‘ƒ(๐ด|๐ธ_1))/(๐‘ƒ(๐ธ_1 ).๐‘ƒ(๐ด|๐ธ_1)+๐‘ƒ(๐ธ_2 ).๐‘ƒ(๐ด|๐ธ_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 = ๐Ÿ—๐ŸŽ% ร— ๐Ÿ—๐ŸŽ% = 90/100 ร— 90/100 = 0.9 ร— 0.9 = 0.81 "P(E2)" = Probability that machine is incorrectly setup = (100โˆ’80)% = 20%=20/100 = 0.2 P(A|E1) = Probability that machine produce 2 acceptable items if it have incorrect set up = ๐Ÿ’๐ŸŽ% ร— ๐Ÿ’๐ŸŽ% = 40/100ร—40/100 = 0.4 ร— 0.4 = 0.16 Putting values in the formula P(E1|A) = (๐‘ƒ(๐ธ_1 ).๐‘ƒ(๐ด|๐ธ_1))/(๐‘ƒ(๐ธ_1 ).๐‘ƒ(๐ด|๐ธ_1)+๐‘ƒ(๐ธ_2 ).๐‘ƒ(๐ด|๐ธ_2) ) = (0.8 ร— ๐ŸŽ.๐Ÿ–๐Ÿ)/(0.8 ร— ๐ŸŽ.๐Ÿ–๐Ÿ + 0.2 ร— ๐ŸŽ.๐Ÿ๐Ÿ”) = 0.648/0.680 = 0.95

About the Author

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.