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Chapter 13 Class 12 Probability
Example 6
Ex 13.1, 10 (a) Important
Ex 13.1, 12 Important
Example 11 Important
Ex 13.2, 7 Important
Ex 13.2, 11 (i)
Ex 13.2, 14 Important
Example 17 Important
Example 18 Important
Example 20 Important
Example 21 Important
Ex 13.3, 2 Important
Ex 13.3, 4 Important
Ex 13.3, 8 Important You are here
Ex 13.3, 10 Important
Ex 13.3, 12 Important
Ex 13.3, 13 (MCQ) Important
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Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
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Example 23 Important
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Misc 7 Important
Misc 10 Important
Chapter 13 Class 12 Probability
Last updated at May 29, 2023 by Teachoo
Ex 13.3, 8 A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?Let A : Items produced by machine A B : Items produced by machine B D : Items is defective We need to find the Probability that the item was produced by machine B, if it is found to be defective i.e. P(B"|"D) P(B"|"D) = (𝑃(𝐵) . 𝑃(𝐷|𝐵))/(𝑃(𝐵) . 𝑃(𝐷|𝐵)+𝑃(𝐴) . 𝑃(𝐷|𝐴) ) P(A) = Probability that the item is produced by machine A = 60%=60/100=𝟎.𝟔 P(D|A) = Probability that the item is defective, if produced by machine A = 2%=2/100=𝟎.𝟎𝟐 P(B) = Probability that the item is produced by machine B = 40%=40/100=𝟎.𝟒 P(D|B) = Probability that the item is defective, if produced by machine B = 1%=1/100=𝟎.𝟎𝟏 Putting values in formula, "P(B|D)" = (0. 4 × 0. 01)/( 0. 6 × 0. 02 +0. 4 ×0. 01 ) = (0. 01 × 0. 4)/(0. 01 [0.6 × 2 + 0.4] ) = 0.4/(1.2 + 0.4) = 0.4/1.6 = 4/16 = 1/4 Therefore, required probability is 1/4