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Three machines E1, E2, E3 in a certain factory produced 50%, 25% and 25%, respectively, of the total daily output of electric tubes. It is known that 4% of the tubes produced one each of

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प्रश्न

Three machines E1, E2, E3 in a certain factory produced 50%, 25% and 25%, respectively, of the total daily output of electric tubes. It is known that 4% of the tubes produced one each of machines E1 and E2 are defective, and that 5% of those produced on E3 are defective. If one tube is picked up at random from a day’s production, calculate the probability that it is defective.

योग
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उत्तर

Let D be the event that the picked up tube is defective

Let A1, A2 and A3 be the events that the tube is produced on machines E1, E2 and E3, respectively.

P(D) = P(A1) P(D|A1) + P(A2) P(D|A2) + P(A3) P(D|A3) ......(1)

P(A1) = `50/100 = 1/2`

P(A2) = `1/4`

P(A3) = `1/4`

Also P(D|A1) = P(D|A2

= `4/100`

= `1/25`

P(D|A3) = `5/100`

= `1/20`

Putting these values in (1), we get

P(D) = `1/2 xx 1/25 + 1/4 xx 1/25 + 1/4 xx 1/20`

= `1/50 + 1/100 + 1/80`

= `17/400`

= 0.0425

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अध्याय 13: Probability - Solved Examples [पृष्ठ २६४]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 13 Probability
Solved Examples | Q 6 | पृष्ठ २६४

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