Respuesta :
Answer:
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
Answer:
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%
