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weqwewe [10]
4 years ago
11

Two assembly lines I and II have the same rate of defectives in their production of voltage regulators. Six regulators are sampl

ed from each line and tested. Among the total of twelve tested regulators, four are defective. Find the probability that exactly two of the defective regulators came from line I. (Round your answer to four decimal places.)
Mathematics
1 answer:
Kamila [148]4 years ago
4 0

Answer:

0.4546

Step-by-step explanation:

nCr = n!/(n-r)!r!

Number of ways of selecting the four defective voltage regulators from 12 =  12C4 = 12!/(12-4)!4! = 12!/8!4! = (12 *11*10*9)/(4*3*2*1)

12C4 =  495 ways

Number of ways of selecting 2 defectives from line 1 = 6C2 * 6C2

6C2 = 6!/(6-2)!2! = 6!/4!2! = (6*5)/(2*1) = 15

6C2 * 6C2 = 15*15 = 225 ways

Probability = Number of possible outcomes/ Number of total outcomes

Probability that exactly 2 of the defective regulators came from line 1 = 225/40.95 = 0.4546

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Answer: 0.99822

Step-by-step explanation:

Our inter-arrival time follows Poisson distribution with parameter 14 which is in hour so first we have to calculate this in minutes as we have to calculate probability in minutes.

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Let X=inter-arrival time between two customers

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Probability(X is less than or equal to 2 minutes) = P(X=0) + P(X=1) + P(X=2)

Now the Poisson distribution has PDF = \frac{\exp ^{-\lambda }\times \lambda ^{X}}{X!}

So P(X = 0) = \frac{\exp ^{-0.233}\times 0.233^{0}}{0!} = 0.79215

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Now adding all three probability gives =  0.79215 +0.18457 + 0.02150=0.99822

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3 years ago
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Robert collected 330 aluminum cans. If 100 aluminum cans weigh 4 pounds, and the recycling center is paying 25 cents a pound, ho
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First, determine the weight of 330 cans with the given conditions,
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