- 1). Draw 10 light bulbs from an infinite pool with a 0.5 percent rate of defects. Suppose that you want to calculate the probability that exactly two bulbs that you drew are defective.
- 2). Calculate the probability that the first bulb drawn is defective. That equals 0.005.
- 3). Calculate the probability that the second bulb drawn is defective. That equals 0.005.
- 4). Calculate the probability that the first two bulbs drawn are defective. This is 0.005^2, or 2.5x10^-5 (in scientific notation, where the caret ^ indicates exponentiation). The reason you multiply them is that you treat the two drawings as a single event---the event being your drawing the 1- bulbs and getting one outcome (two defectives).
- 5). Calculate the probability that the third through 10th bulb drawn is not defective. That is 0.995^8 = 0 .960693. Again, you multiply the drawings within a single event.
- 6). Calculate the number of ways the 10 bulbs could be arranged. For example, bulb 1 could have been drawn third and bulb 10 could have been drawn second, and so on. Therefore, you will multiply the probabilities from Steps 4 and 5 by 10!, where the factorial sign, !, means 10x9x8x7x6x5x4x3x2x1.
- 7). Calculate the number of orderings of the two defective bulbs, since your original question doesn't care about the ordering of those bulbs. So you'll eventually divide out 2!.
- 8). Calculate the number of orderings of the eight functional bulbs. Following the above logic, this equals 8!.
- 9). Calculate the result of 10!/2!8! to be 45. Multiply this number by the results of Steps 4 and 5 to get 45 x (2.5x10^-5) x 0.960693 = 1.081x10^-3. This is the probability of getting precisely two defective bulbs out of a sample of 10.
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