Question

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

Number of defective bulbs	0	1	2	3	4	5	6	more than 6
Frequency	400	180	48	41	18	8	3	2

One carton was selected at random. What is the probability that it has

1. no defective bulb?
2. defective bulbs from 2 to 6?
3. defective bulbs less than 4?

Answer 1

Total number of cartons, n(S) = 700

i. Number of cartons which has no defective bulb,

 n(E₁) = 400

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

= `400/700`

= `4/7`

Hence, the probability that no defective bulb is `4/7`.

ii. Number of cartons which has defective bulbs from 2 to 6, 

n(E₂) =  48 + 41 + 18 + 8 + 3 = 118

∴ Probability that the defective bulbs from 2 to 6 = `(n(E_2))/(n(S))`

= `118/700`

= `59/350`

Hence, the probability that the defective bulbs from 2 to 6 is `59/350.`

iii. Number of cartons which has defective bulbs less than 4,

n(E₃) =  400 + 180 + 48 + 41 = 669

∴ The probability that the defective bulbs less than 4 = `(n(E_3))/(n(S)) = 669/700`

Hence, the probability that the defective bulbs less than 4 is `669/700.`