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
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).
To find the probability of choosing at least three lamps from a pack of 12 where 4 are defective and the rest are non-defective, we need to consider two scenarios: choosing exactly three lamps and choosing all four defective lamps.
Scenario 1: Choosing exactly three lamps
The number of ways to choose 3 lamps from a pack of 12 is given by the combination formula: C(12, 3) = 12! / (3! * (12-3)!) = 220.
Out of these 220 combinations, we need to ensure that at least 2 out of the 3 lamps chosen are non-defective, as per the requirement. The number of ways to choose 2 non-defective and 1 defective lamp is C(8, 2) * C(4, 1) = 28 * 4 = 112.
So, the probability of choosing exactly three lamps with at least two non-defective ones is 112/220 = 28/55.
Scenario 2: Choosing all four defective lamps
The number of ways to choose 4 lamps from a pack of 12 is given by the combination formula: C(12, 4) = 12! / (4! * (12-4)!) = 495.
However, if we choose all four defective lamps, then there are no non-defective lamps left to choose. Therefore, this scenario does not fulfill the requirement of choosing at least three lamps.
So, the probability of choosing all four defective lamps is 0.
Finally, we need to find the probability of either scenario happening, which is the sum of their probabilities:
Probability = Probability of choosing exactly three lamps + Probability of choosing all four defective lamps
Probability = 28/55 + 0 = 28/55