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Helga [31]
3 years ago
11

Suppose a shipment of 190 electronic components contains 3 defective components. To determine whether the shipment should be​ ac

cepted, a​ quality-control engineer randomly selects 3 of the components and tests them. If 1 or more of the components is​ defective, the shipment is rejected. What is the probability that the shipment is​ rejected?
Mathematics
1 answer:
PolarNik [594]3 years ago
5 0

Answer:

0.0467

Step-by-step explanation:

Probabilty of being defective is  \frac{3}{190}

So probability of being good is  \frac{187}{190}

Now,

Probability that all 3 of them are good is:

\frac{187}{190}^{3}=0.9533

Now, to find probability of rejection, we subtract the now found probability from 1. Thus, we have:

1 - 0.9533 = 0.0467

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Answer:

x = 20/13 , y = 16/13 , z = 120/13

Step-by-step explanation:

Solve the following system:

{x + y + z = 12 | (equation 1)

6 x - 2 y + z = 16 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Swap equation 1 with equation 2:

{6 x - 2 y + z = 16 | (equation 1)

x + y + z = 12 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Subtract 1/6 × (equation 1) from equation 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+(4 y)/3 + (5 z)/6 = 28/3 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Multiply equation 2 by 6:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

3 x + 4 y + 2 z = 28 | (equation 3)

Subtract 1/2 × (equation 1) from equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

0 x+5 y + (3 z)/2 = 20 | (equation 3)

Multiply equation 3 by 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+8 y + 5 z = 56 | (equation 2)

0 x+10 y + 3 z = 40 | (equation 3)

Swap equation 2 with equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+8 y + 5 z = 56 | (equation 3)

Subtract 4/5 × (equation 2) from equation 3:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+(13 z)/5 = 24 | (equation 3)

Multiply equation 3 by 5:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+13 z = 120 | (equation 3)

Divide equation 3 by 13:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y + 3 z = 40 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Subtract 3 × (equation 3) from equation 2:

{6 x - 2 y + z = 16 | (equation 1)

0 x+10 y+0 z = 160/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Divide equation 2 by 10:

{6 x - 2 y + z = 16 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Add 2 × (equation 2) to equation 1:

{6 x + 0 y+z = 240/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Subtract equation 3 from equation 1:

{6 x+0 y+0 z = 120/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Divide equation 1 by 6:

{x+0 y+0 z = 20/13 | (equation 1)

0 x+y+0 z = 16/13 | (equation 2)

0 x+0 y+z = 120/13 | (equation 3)

Collect results:

Answer:  {x = 20/13 , y = 16/13 , z = 120/13

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