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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
- no defective bulb?
- defective bulbs from 2 to 6?
- defective bulbs less than 4?
Solution
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.`
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