Question

Over the past 200 working days, the number of defective parts produced by a machine is given in the following table:

Number of defective parts	0	1	2	3	4	5	6	7	8	9	10	11	12	13
Days	50	32	22	18	12	12	10	10	10	8	6	6	2	2

Determine the probability that tomorrow’s output will have not more than 5 defective parts

Answer 1

Total number of working days, n(S) = 200

Number of days in which not more than 5 defective parts,

n(E₃) = 50 + 32 + 22 + 18 + 12 + 12 = 146

∴ Probability that not more than 5 defective parts

= `(n(E_3))/(n(S)) = 146/200` = 0.73