Two production lines are used to pack sugar into 5 kg bags. Line 1 produces twice as many bags as does line 2. One percent of bags from line 1 are defective in that they fail to meet a purity specification, while 3% of the bags from line 2 are defective. A bag is randomly chosen for inspection. ( 2 points, each part is 0.5 points) b. What is the probability that is defective

P(Bag is Defective) = 0.0167

Step-by-step explanation:

Line 1 produces twice as many bags as line 2. Let x be the number of bags produced by line 2.

No. of bags produced by line 2 = x

No. of bags produced by line 1 = 2x

Probability that the bag has been produced by line 1 can be written as:

P(Line 1) = No. of bags produced by line 1/Total no. of bags

             = 2x/(x+2x)

             = 2x/3x

P(Line 1) = 2/3. Similarly,

P(Line 2) = x/3x

P(Line 2) = 1/3

1% bags produced by line 1 are defective so the probability of line 1 producing a defective bag is:

P(Defective|Line 1) = 0.01

3% of bags from line 2 are defective, so:

P(Defective|Line 2) = 0.03

b. The probability that the chosen bag is defective can be calculated through the conditional probability formula:

P(A|B) = P(A∩B)/P(B)

P(A∩B) = P(A|B)*P(B)

The chosen defective bag can be either from line 1 or from line 2. So, the probability that the chosen bag is defective is:

P(Bag is Defective) = P(Defective and from Line 1) + P(Defective and from Line 2)

                                = P(D∩Line 1) + P(D∩Line 2)

                                = P(Defective|Line 1)*P(Line 1) + P(Defective|Line 2)*P(Line 2)

                                = (0.01)*(2/3) + (0.03)(1/3)

P(Bag is Defective) = 0.0167

P(z) = 5/300 = 0.0167 or 1.67%

The probability that a bag randomly chosen for inspection is defective is 0.0167

Step-by-step explanation:

Let x, y and z represent the number of bags of sugar produced by line 1, line 2 and both respectively.

z = x + y 1

Line 1 produces twice as many bags as does line 2

x = 2y 2

Substituting into equation 1

z = 2y + y

z = 3y  3

One percent of bags from line 1 are defective

N(x) = 0.01x= 0.01(2y)           (x = 2y)

N(x)  = 0.02y          

3% of the bags from line 2 are defective.

N(y) = 0.03y

The total number of defective bags from both lines

N(z) = 0.02y + 0.03y = 0.05y 4

The probability that a bag randomly chosen for inspection is defective;

P(z) = N(z)/z  5

Substituting equation 3 and 4

P(z) = 0.05y/3y

P(z) = 5/300 = 0.0167 or 1.67%

it’s supposed to be -4+45=41

step-by-step explanation: