Question 1:

A box of 600 bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. Then the probability that it is non-defective bulb is:

A. 143/150

B. 147/150

Feedback P (non-defective bulb) = 1 – P (Defective bulb)
= 1 – (12/600)
= (600 – 12)/600
147/150

C. 1/25

D. 1/50

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