Bulbs are packed in cartons , each containing 40 bulbs . 700 cartons were examined for defective bulbs and the results are given in the following table :

One carton is selected at random. What is the probability that it has
(i) no defective bulb?
(ii) defective bulbs less than 4 ?
(iii) defective bulbs more than 3 but less than 6 ?
(iv) defective bulbs 6 or more ?

Explanation

Total number of cartons = 700 .
(i) Let `E_(1)` be the event of choosing a carton having no defective bulb . Then ,
P (choosing a carton having no defective bulb)
`= P(E_(1))`
= `("number of cartons having 0 defective bulb")/("total number of cartons")`
`= (371)/(700) = (53)/(100) = 0.53`.
(ii) Let `E_(2)` be the event of choosing a carton having defective bulbs less than 4 . then,
P (choosing a carton having defective bulbs less than 4)
`= P(E_(2))`
= `("number of cartons having defective bulbs 0 , 1 , 2 or 3")/("total number of cartons")`
`(371 + 162 + 55 + 49)/(700) = (637)/(700) = (91)/(100) = 0.91`.
Let `E_(3)` be the event of choosing a carton having defective bulbs more than 3 but less than 6 . Then ,
P (choosing a carton having defective bulbs more than 3 , but less than 6)
`= P(E_(3))`
`("number of cartons having defective bulbs 4 or 5")/("total number of cartons")`
= `(41 + 15)/(700) = (56)/(700) = (8)/(100) = 0.08` .
(iv) Let `E_(4)` be the event of choosing a carton having defective bulbs 6 or more . then ,
P(choosing a carton having defective bulbs 6 or more )
`P (E_(4))`
= `("number of cartons having defective bulbs 6 or more")/("total number of cartons")`
`= (5 + 2)/(700) = (7)/(700) = (1)/(100) = 0.01.`