two lamps are to be choosen from a pack of 12 lamps where 4 are def ective and the rest are non defective what is the probability of at least three lamps

Answer 1

There are different ways to interpret "at least three lamps," so I will provide two possible answers.

1. If you mean "at least three lamps are non-defective," then you can use the complement rule and count the probability of the opposite event, which is "two or fewer lamps are non-defective":

- The total number of ways to choose two lamps out of 12 is C(12,2) = 66.
- The number of ways to choose two non-defective lamps out of 8 (since there are 4 defective lamps) is C(8,2) = 28.
- The number of ways to choose one non-defective lamp and one defective lamp is C(8,1) x C(4,1) = 32.
- The number of ways to choose two defective lamps out of 4 is C(4,2) = 6.
- Therefore, the number of ways to choose two or fewer non-defective lamps is 28 + 32 + 6 = 66.
- Hence, the probability of choosing at least three non-defective lamps is 1 - (66/ C(12,2)) = 0.5455 (rounded to 4 decimal places).

2. If you mean "at least three lamps are selected," regardless of their defectiveness, then you can compute the probabilities of selecting exactly 3, 4, or 5 lamps, and add them up:

- The total number of ways to choose three lamps out of 12 is C(12,3) = 220.
- The number of ways to choose exactly 3 non-defective lamps is C(8,3) = 56, since you need to choose 3 out of the 8 non-defective lamps.
- The number of ways to choose 2 non-defective and 1 defective lamp is C(8,2) x C(4,1) = 112, since you need to choose 2 out of the 8 non-defective lamps and 1 out of the 4 defective lamps.
- The number of ways to choose 1 non-defective and 2 defective lamps is C(8,1) x C(4,2) = 42, since you need to choose 1 out of the 8 non-defective lamps and 2 out of the 4 defective lamps.
- Therefore, the probability of choosing exactly 3 lamps is (56 + 112 + 42) / 220 = 0.7636 (rounded to 4 decimal places).
- Similarly, the total number of ways to choose four lamps out of 12 is C(12,4) = 495, and you can compute the number of ways to choose exactly 4 non-defective lamps, or 3 non-defective and 1 defective, or 2 non-defective and 2 defective, or 1 non-defective and 3 defective, which gives a probability of (70 + 224 + 132 + 15) / 495 = 0.6657.
- The total number of ways to choose five lamps out of 12 is C(12,5) = 792, and you can compute the number of ways to choose exactly 5 non-defective lamps, or 4 non-defective and 1 defective, or 3 non-defective and 2 defective, which gives a probability of (56 + 224 + 192) / 792 = 0.6061.
- Therefore, the probability of choosing at least three lamps is the sum of the probabilities of choosing exactly 3, 4, or 5 lamps, which is 0.7636 + 0.6657 + 0.6061 = 2.0354 (rounded to 4 decimal places). However, this probability is greater than 1, which is impossible, so we must have made a mistake. The mistake is that we counted some outcomes multiple times, since the events of choosing 3, 4, or 5 lamps are not mutually exclusive. For example, the outcome of choosing 3 non-defective lamps is also counted in the outcome of choosing 4 lamps, since one of the 4 lamps could be defective. Therefore, we need to subtract the probabilities of the intersections of these events to avoid double-counting. For example, the probability of choosing exactly 4 lamps that include 3 non-defective lamps is equal to the number of ways to choose 3 non-defective lamps out of 8 times the number of ways to choose 1 lamp out of 4, which is C(8,3) x C(4,1) = 112, as we calculated before. Similarly, the probability of choosing exactly 5 lamps that include 3 or 4 non-defective lamps is equal to the number of ways to choose 3 non-defective lamps out of 8 times the number of ways to choose 2 lamps out of 4, plus the number of ways to choose 4 non-defective lamps out of 8 times the number of ways to choose 1 lamp out of 4, which is C(8,3) x C(4,2) + C(8,4) x C(4,1) = 132 + 70 = 202. Therefore, the corrected probability of choosing at least three lamps is 0.7636 + (0.6657 - 0.2244) + (0.6061 - 0.1313) = 1.8797 (rounded to 4 decimal places).