Question

In a sample of 100 logic chips, 23 have a defect DI, 26 have a defect D2. 30
have a defect D3, 7 have defect Dl and D2, & have defect Di and D3. 10
have defect D2 and D3 and 3 have all the three defects. Find the
number of chips huving
(i) at Icast one defect (ii) no defect.​

Answer 1

Step-by-step explanation:

Annexure contains solution.

Answer 2

Venn Diagram

Given:

There are 100 logic chips.

23 chips have defect D1.

26 chips have defect D2.

30 chips have defect D3.

7 chips have defect D1 and defect D2.

8 chips have defect D1 and defect D3.

7 chips have defect D2 and defect D3.

3 chips have defect D1, D2 and D3.

To find:

Number of chips having at least one defect and no defect respectively.

Explanation:

This type of problems can be solved using Venn diagram.

It is an illustration that uses circles to show the relationships among things or finite groups of things.

An image containing required Venn diagram is attached,

(i) Hence, in finding the number of chips having at least one defect will be = $n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\capA)+n(A\cap B\cap C)\\\\23+26+30-7-8-10+3=57$

Hence, 57 chips has at least one defect.

(ii) Number of chips having no defect= (Total number of chips) - (Number of chips having at least one defect) =$100-57=43$,

So, 57 chips are there having at least one defect and 43 chips have no defect.