Question

assume that the number of production defects is 28 and that 14 of these are classified as major defects, 8 are classified as design defects, and 11 were neither major defects nor design defects. How many of the design defects were major?

Answer 1

Answer:  5

Step-by-step explanation:

Let A denotes the number of major defects and B denotes the number of design defect.

By considering the given information, we have

$U=28\ ;\ n(A)=14\ ;\ n(B)=8\ \ ;\ n(A^c\cap B^c)=1$

Now, the number of major defects or design defects:

$n(A\cup B)=U-n(A^c\cap B^c)=28-11=17$

Also,

$n(A\cap B)=n(A)+n(B)-n(A\cup B)\\\\\Rightarrow\ n(A\cap B)=14+8-17=5$

Hence, the number of  design defects were major=5