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

From these days , one day is chosen at random . What is the probability that on that day , the output has
(i) no defective part ?
(ii) at least 1 defective part ?
(iii) not more than 5 defective parts ?
(iv) more than 5 , but less than 8 defective parts ?
(v) more than 13 defective parts ?

Explanation

Total number of working days = 200.
(i) Let `E_(1)` be the event that the output has 0 defective part on the chosen day . Then ,
P(event of producing 0 defective part on the chosen day)
`= P(E_(1))`
`("number of days when the output has 0 defective part")/("total number of working days ")`
`= (50)/(200) = (1)/(4) = 0.25` .
(ii) Number of days on which the output has at least 1 defective part
= 200 - number of days with 0 defective part
= 200 - 50 = 150 .
Let `E_(2)` be the event that the output has at least 1 defective part on the chosen day . Then ,
`P(E_(2)) = (150)/(200) = (3)/(4) `.
(iii) Let `E_(3)` be the event that the output has not more than 5 defective parts , i.e., 5 or less defective parts , on the chosen day . Then ,
P(events that the output has not more than 5 defective parts on the chosen day)
= P (event that the output has 5 or less defective parts on the chosen day)
= `P(E_(3))`
`= ("number of days when the output has 5 or less defective parts")/("total number of working days")`
`= (50 + 32 + 22+ 18 + 12 + 12)/(200) = (146)/(200) = (73)/(100) = 0.73`.
(iv) Let `E_(4)` be the event that the output has more than 5 , but less than 8 defective parts on the chosen day . Then ,
P (event that the output has more than 5 but less than 8 defective parts on the chosen day)
`= P(E_(4))`
=`("number of days when the output has 6 or 7 defective parts")/("total number of working days")`
`= (10 + 10)/(200) = (20)/(200) = (1)/(10) = 0.1`.
(v) Let `E_(5)` be the event that the output has more than 13 defective parts on that day . then ,
P (event that the output has more than 13 defective parts on that day)
`= P(E_(5))`
`= ("number of days when the output has more than 13 defective parts")/("total number of working days")`
=`(0)/(200) = 0.`