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Example 37 - If a machine is correctly set up, it produces 90%

Example 37 - Chapter 13 Class 12 Probability - Part 2
Example 37 - Chapter 13 Class 12 Probability - Part 3

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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) = (𝑃(𝐸_1 ).𝑃(𝐴|𝐸_1))/(𝑃(𝐸_1 ).𝑃(𝐴|𝐸_1)+𝑃(𝐸_2 ).𝑃(𝐴|𝐸_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 = πŸ—πŸŽ% Γ— πŸ—πŸŽ% = 90/100 Γ— 90/100 = 0.9 Γ— 0.9 = 0.81 "P(E2)" = Probability that machine is incorrectly setup = (100βˆ’80)% = 20%=20/100 = 0.2 P(A|E1) = Probability that machine produce 2 acceptable items if it have incorrect set up = πŸ’πŸŽ% Γ— πŸ’πŸŽ% = 40/100Γ—40/100 = 0.4 Γ— 0.4 = 0.16 Putting values in the formula P(E1|A) = (𝑃(𝐸_1 ).𝑃(𝐴|𝐸_1))/(𝑃(𝐸_1 ).𝑃(𝐴|𝐸_1)+𝑃(𝐸_2 ).𝑃(𝐴|𝐸_2) ) = (0.8 Γ— 𝟎.πŸ–πŸ)/(0.8 Γ— 𝟎.πŸ–πŸ + 0.2 Γ— 𝟎.πŸπŸ”) = 0.648/0.680 = 0.95

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.