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A VCR manufacturer receives 70% of his parts from factory F1 and the rest from factory F2. Suppose that 3% of the output from F1 are defective while only 2% of the output from F2 are defective. a. What is the probability that a received part is defective? b. If a randomly chosen part is defective, what is the probability it came from factory F1?

Respuesta :

Answer:

See the explanation.

Step-by-step explanation:

Lets take 1000 output in total.

From these 1000 outputs, 700 are from F1 and 300 are from F2.

Defective from F1 is [tex]\frac{700\times3}{100} = 21[/tex] and defective from F2 is [tex]\frac{300\times2}{100} = 6[/tex].

a.

Total received part is 1000.

Total defectives are (21+6) = 27.

The probability of a received part being defective is [tex]\frac{27}{1000}[/tex].

b.

The probability of the randomly chosen defective part from F1 is [tex]\frac{21}{27} = \frac{7}{9}[/tex].

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