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%

the solution is x=1

step-by-step explanation:

we are given our   equation as

let's assume

left side function is

right side function is

now, we can draw graph

and then we can get intersection point

the intersection point will be solution

so, the solution is

x=1

b. b < 1 might be the correct answer