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 more than 13 defective parts

Answer 1

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

Number of days in which more than 13 defective parts, n(E₄) = 0

= `(n(E_4))/(n(S)) = 0/200` = 0

Hence, the probability that more than 13 defective parts is 0.