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There are 3 bulbs in the box and two are defective so on the first draw you have a 2/3 probability of selecting a defective bulb and a 1/3 probability of drawing a bulb that is not defective. If you draw a defective bulb then there are 2 bulbs remaining, one defective and one not defective. Thus on the second draw the probability of selecting a defective bulb is 1/2 and the probability of drawing a nondefective bulb is 1/2. If your first draw was a nondefective bulb then the 2 remaining bulbs are both defective and hence the probability of drawing a defective bulb on the second draw is 1. If you draw two bulbs then you take one of the three paths through the tree. Since 2/3 × 1/2 = 1/3 and 1/3 × 1 = 1/3, each of the three paths has a probability of 1/3.
If you chose a nondefective and a defective bulb you did so by either choosing a nondefective and then a defective or a defective and then a nondefective. Each possibility has probability 1/3 so the probability of choosing a nondefective and a defective bulb is 1/3 + 1/3 = 2/3.
Check to see if this question is correctly worded. If it is then the probability is 0 since you can't draw 3 defective bulbs, there are only 2.
Extend the tree diagram another level to answer this question. Penny  


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