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

One carton was selected at random. What is the probability that it has
i) no defective bulb?
ii) defective bulbs from 2-6?
iii) defective bulbs less than 4?

Explanation

Total numbers of cartons, n(S)=700
i) Number of cartons which has no defective bulb, `n(E_(1))=400`
`therefore` Probability that no defective bulb`=(n(E_(1)))/(n(S))=400/700=4/7`
Hence, the probability that no defective bulbs is `4/7`.
ii) Number of cartons which has defective bulbs from 2 to 6, `n(E_(2))`
`=48+41+18+8+3=118`
`therefore` Probability that the defective bulbs from 2 to 6 `=(n(E_(2)))/(n(S))=118/70=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(E_(3)) = 400+180+48+41=669`
`therefore` The probability that the defective bulbs less than 4`=(n(E_(3)))/(n(S))=669/700`
Hence, the probability that the defective bulb less than 4 is `669/700`.