Example 37 - Chapter 13 Class 12 Probability (Term 2)
Last updated at Feb. 15, 2020 by Teachoo
Bayes theorem
Ex 13.3, 2 Important
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Misc 3
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Ex 13.3, 13 (MCQ) Important
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Example 18 Important
Example 17 Important
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Example 37 Important You are here
Example 20 Important
Example 33 Important
Misc 12
Misc 16 Important
Misc 13 Important
Bayes theorem
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) = (π(πΈ_1 ).π(π΄|πΈ_1))/(π(πΈ_1 ).π(π΄|πΈ_1)+π(πΈ_2 ).π(π΄|πΈ_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 = ππ% Γ ππ% = 90/100 Γ 90/100 = 0.9 Γ 0.9 = 0.81 "P(E2)" = Probability that machine is incorrectly setup = (100β80)% = 20%=20/100 = 0.2 P(A|E1) = Probability that machine produce 2 acceptable items if it have incorrect set up = ππ% Γ ππ% = 40/100Γ40/100 = 0.4 Γ 0.4 = 0.16 Putting values in the formula P(E1|A) = (π(πΈ_1 ).π(π΄|πΈ_1))/(π(πΈ_1 ).π(π΄|πΈ_1)+π(πΈ_2 ).π(π΄|πΈ_2) ) = (0.8 Γ π.ππ)/(0.8 Γ π.ππ + 0.2 Γ π.ππ) = 0.648/0.680 = 0.95