InterviewSolution
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In a bolt factory, machines A, B, and C manufacture respectively 25%, 35% and 40% of the total bolts. Of their output 5%, 4% and 2% are respectively defective bolt. On the basis of the above information answer the following, If drawn bolt is defective , then the probability that it is not manufactured by C is(a) \(\frac{39}{69}\) (b) \(\frac{40}{69}\) (c) \(\frac{41}{69}\) (d) \(\frac{42}{69}\)My answer is coming to be \(\frac{53}{69} \), which is not in the options. Please give CBSE Boards oriented solution with steps. |
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Answer» Let events A,B and C are A : Bolt is manufactured by machine A. B : Bolt is manufactured by machine B. C : Bolt is manufactured by machine C. P(A) = 25% = \(\frac{25}{100}\) = \(\frac{1}{4}\) P(B) = 35% = \(\frac{35}{100} = \frac{7}{20}\) P(C) = 40% = \(\frac{40}{100} = \frac{2}{5}\) let E be event that drawn bold in defective. P(E/A) = 5% = \(\frac{5}{100} = \frac{1}{20}\) P(E/B) = 4% = \(\frac{4}{100} = \frac{1}{25}\) P(E/C) = 2% = \(\frac{2}{100} = \frac{1}{20}.\) Probability that defective bolt is manufactured by C is P(C/E) = \(\frac{P(E/C)P(C)}{P(E/C).P(C)+P(E/B)\, P(B)+P(E/A)P(A)}\) = \(\cfrac{\frac{1}{50} \times \frac{2}{5}}{\frac{1}{50}\times \frac{2}{5} + \frac{1}{25} \times \frac{7}{20}+ \frac{1}{20}\times\frac{1}{4}}\) =\(\cfrac{\frac{1}{125}}{\frac{1}{125} + \frac{7}{500}+\frac{1}{80}}\) = \(\cfrac{\frac{1}{125}}{\frac{16+28+25}{2000}}\) = \(\frac{16}{69}\) Thus probability that defective bolt is not manufactured by C is P(C'/E) = 1-P(C/E) = 1 - \(\frac{16}{69}\) = \(\frac{69-16}{69}\) = \(\frac{53}{69}\). |
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