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Example 19 In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?Let A : bolt manufactured from machine A B : bolt manufactured from machine B C : bolt manufactured from machine C D : bolt is defective We need to find the Probability that the bolt is manufactured by machine B, if it is defective i.e. P(B"|"D) So, "P(B|D) = " (𝑃(𝐵). 𝑃(𝐷|𝐵))/(𝑃(𝐴). 𝑃(𝐷|𝐴) + 𝑃(𝐵). 𝑃(𝐷|𝐵) + 𝑃(𝐶). 𝑃(𝐷|𝐶) ) "P(A)" = Probability that the bolt is made by machine A = 25%=25/100=𝟎.𝟐𝟓 𝑷(𝐃"|A") = Probability of a defective bolt from machine A = 5%=5/100=𝟎.𝟎𝟓 "P(B)" = Probability that the bolt is made by machine B = 35%=35/100=0.35 𝑷(𝐃"|B") = Probability of a defective bolt from machine B = 4%=4/100=𝟎.𝟎𝟒 "P(C)" = Probability that the bolt is made by machine C = 40%=40/100=0.40 𝑷(𝐃"|C") = Probability of a defective bolt from machine C = 2%=2/100=𝟎.𝟎𝟐 Putting Values in formula, "P(B|D) = " (𝟎.𝟑𝟓 × 𝟎.𝟎𝟒)/(𝟎.𝟐𝟓 × 𝟎.𝟎𝟓 + 𝟎.𝟑𝟓 × 𝟎.𝟎𝟒 + 𝟎.𝟒𝟎 × 𝟎.𝟎𝟐) = 0.014/( 0.0125 + 0.014 + 0.008 ) = 0.014/( 0.0345 ) = 140/( 345 ) = 28/69 Therefore, required probability is 𝟐𝟖/𝟔𝟗

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo