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jaycobgarciavis
05.05.2020 •
Mathematics
A chip company has two manufacturing plants. Plant A produces 40% of the chips and Plant B produces 60% of the chips produced by the company. The company knows that 2% of the chips produced by plant A are defective and 1% of the chips produced by plant B are defective. If a randomly chosen chip produced by the company is defective, what is the likelihood that the chip came from plant A
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Ответ:
P(A/D) = 0.5714
Step-by-step explanation:
Let's call A the event that a chip is produced by Plant A, B the event that a chip is produced by Plant B and D the event that the chip is defective
So, the likelihood or probability P(A/D) that a chip came from plant A given that the chip is defective is calculated as:
P(A/D) = P(A∩D)/P(D)
Where P(D) = P(A∩D) + P(B∩D)
Then, the probability P(A∩D) that a chip is produced by plant A and it is defective is calculated as:
P(A∩D) = 0.4*0.02 = 0.008
Because, Plant A produces 40% of the chips and 2% of the chips produced by plant A are defective.
At the same way, the probability P(B∩D) that a chip is produced by plant B and it is defective is calculated as:
P(B∩D) = 0.6*0.01 = 0.006
So, P(D) and P(A/D) are equal to:
P(D) = 0.008 + 0.006 = 0.014
P(A/D) = 0.008/0.014 = 0.5714
it means that if a randomly chosen chip produced by the company is defective, the likelihood that the chip came from plant A is 0.5714
Ответ:
answer:
-3.25
step-by-step explanation:
[ 3/4*-4] = -3
[-2/3*2] = -1.3
-3 x -1.3 = 3.9
[3/5]*-2 = -1.2
3.9 ÷ -1.2 = -3.25