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

1. no defective part
2. atleast one defective part
3. not more than 5 defective parts
4. more than 13 defective parts

Answer 1

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

i. Number of days in which no defective part is,

n(E₁) = 50

Probability that no defective part = `(n(E_1))/(n(S))`

= `50/200`

= `1/4`

= 0.25

ii. Number of days in which atleast one defective part is,

n(E₂) = 32 + 22 + 18 + 12 + 12 + 10 + 10 + 10 + 8 + 6 + 6 + 2 + 2 = 150

∴ Probability that atleast one defective part = `(n(E_2))/(n(S))`

= `150/200`

= `3/4`

= 0.75

iii. 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

iv. 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.