Certain pieces made by an automatic lathe are subject to three kinds of defects X, Y, Z. A sample of 1000 pieces was inspected with the following results: 2.1% had type X defect. 24% had type Y defect. 2.8% had type Z defect. 0.3% had both type X and type Y defects. 04% had both type X and type Z defects. 0.6% had both type Y and type Z defects. 0.1% had type X, type Y, and type Z defects. Draw a Venn Diagram, then find:
(a) What percent had none of these defects?
(b) What percent had at least one of these defects?
(c) What percent were free of type X and/or type Y defects?
(d) What percent had not more than one of these defects?

Number of samples containing none of these defects = 1000 - { (Type X) + (Type Y) + (Type Z) - (Both X and Y) - (Both X and Z) - (Both Y and Z) + (All defects X and Y and Z) }

Number of samples containing none of these defects = 1000 - { (21) + (24) +(28) - (3) -(4) - (6) + (1) }

Number of samples containing none of these defects = 1000 - 61

Number of samples containing none of these defects = 939

% age of samples containing none of these defects = 939/1000 x 100

% age of samples containing none of these defects = 93.9%

b) % age of samples containing at least one of these defects:

We have already calculated this above, number of samples containing at least on of these defects:

number of samples containing at least on of these defects = { (Type X) + (Type Y) + (Type Z) - (Both X and Y) - (Both X and Z) - (Both Y and Z) + (All defects X and Y and Z) }

number of samples containing at least on of these defects = { (21) + (24) +(28) - (3) -(4) - (6) + (1) }

number of samples containing at least on of these defects = 61

% age of samples containing at least one of these defects = 61/1000 x 100

% age of samples containing at least one of these defects = 6.1%

c) % age of samples free of type X and/or type Y defects.

For this find, we need to find the samples with only Z Type defect.

Number of Samples with Only Z type defects = { (Type Z) - (Both X and Z) - (Both Y and Z) + (All defects X and Y and Z) }

Number of Samples with Only Z type defects = { (28) -(4) - (6) + (1) }

Number of Samples with Only Z type defects = 19

Now, we also know the number of samples without any defects = 939

Now,

The number of samples free of type X and/or type Y defect = Sum of Number of Samples with Only Z type defects and number of samples without any defects

The number of samples free of type X and/or type Y defect = 19+939

The number of samples free of type X and/or type Y defect = 958

% age of samples free of type X and/or type Y defects = 958/1000 x 100

% age of samples free of type X and/or type Y defects = 95.8%

d) %age of samples with not more than 1 defect:

For this find, we need to find number of samples with only X type and with only type Y and with only type Z.

We have already found the number of samples with only Z type defect = 19

Now,

number of samples with only X type defect = { (Type X) - (Both X and Z) - (Both X and Y) + (All defects X and Y and Z) }

number of samples with only X type defect = { (21) -(4) - (3) + (1) }

number of samples with only X type defect = 15

Similarly,

number of samples with only Y type defect = { (Type Y) - (Both Y and Z) - (Both X and Y) + (All defects X and Y and Z) }

number of samples with only Y type defect = { (24) -(6) - (3) + (1) }

number of samples with only Y type defect = 16

For,

samples with not more than 1 defect = number of samples with only Y type defect + number of samples with only X type defect + number of samples with only z type defect + number of samples without any defects

samples with not more than 1 defect = 939 + 16 + 15 + 19

samples with not more than 1 defect = 989

%age of samples with not more than 1 defect = 989/1000 x 100

%age of samples with not more than 1 defect = 98.9%