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1. An assembler of computer routers and modems uses parts from three sources. Company A supplies 60%
of the parts, Company B supplies 30% of the parts, and Company C supplies the remaining 10% of the
parts. From past experience, the assembler knows that 3% of the parts supplied by Company A are
defective, 5% of the parts supplied by Company B are defective, and 8% of the parts supplied by
Company C are defective. If a part is selected at random, what is the probability it is defective?

Respuesta :

Answer:

0.041

Step-by-step explanation:

Data provided in the question:

P(A) = 0.60

P(B) = 0.30

P(C) = 0.10

P( d/A ) = 0.03

P(d/B) = 0.05

P(d/C) = 0.08

Here,

d = defective

Now,

P( defective )

= P(supplied by A and defective) + P(supplied by B and defective) + P(supplied by C and defective)

= 0.6 × 0.03 + 0.3 × 0.05 + 0.10 × 0.08

= 0.018 + 0.015 + 0.008

= 0.041

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