Engineering
Mathematics
Conditional Probability
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 :
No. 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. The probability of defective bulbs being less than 4 is

89700

432700

600700

669700

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Solution
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P(carton having defective bulbs less than 4)

= 1 − P(carton having defective bulbs greater than or equal to 4)
=118+8+3+2700=131700=669700
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