A carton of 24 bulbs contain 6 defective bulbs. One bulbs is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?
Probability that the first bulb is not defective:Total bulbs = 24Defective bulbs = 6Non-defective bulbs = 24 – 6 = 18Probability of drawing a non-defective bulb = (Number of non-defective bulbs) / (Total bulbs) = 18/24 = 3/4
Probability that the second bulb is defective, given the first bulb is defective and not replaced:After drawing one defective bulb, remaining bulbs = 24 – 1 = 23Remaining defective bulbs = 6 – 1 = 5Probability of drawing a defective bulb = (Number of defective bulbs) / (Remaining bulbs) = 5/23
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