Through a mix-up on the production line, 6 defective refrigerators were shipped out with 44 good ones. if 5 are selected at random, what is the probability that at least 2 of them are defective?
ty in advance!!
To find the probability of at least 2 defective refrigerators out of 5 selected at random, we can use the concept of combination.
First, let's calculate the total number of ways to choose 5 refrigerators out of the total 50 (6 defective + 44 good) refrigerators:
Total number of ways = C(50, 5) = 50! / (5! * (50 - 5)!) = 2,118,760
Next, let's calculate the number of ways to choose 5 refrigerators with exactly 0 or 1 defective refrigerators:
Number of ways to choose 5 with exactly 0 defects = C(44, 5) = 44! / (5! * (44 - 5)!) = 1,086,008
Number of ways to choose 5 with exactly 1 defect = C(44, 4) * C(6, 1) = (44! / (4! * (44 - 4)!)) * (6! / (1! * (6 - 1)!)) = 98,280
Therefore, the number of ways to choose 5 refrigerators with at least 2 defects is:
Number of ways to choose 5 with at least 2 defects = Total number of ways - Number of ways to choose 5 with exactly 0 defects - Number of ways to choose 5 with exactly 1 defect
= 2,118,760 - 1,086,008 - 98,280
= 934,472
Finally, we can calculate the probability by dividing the number of ways to choose 5 refrigerators with at least 2 defects by the total number of ways to choose 5 refrigerators:
Probability = Number of ways to choose 5 with at least 2 defects / Total number of ways
= 934,472 / 2,118,760
≈ 0.44
Therefore, the probability that at least 2 of the 5 randomly selected refrigerators are defective is approximately 0.44 or 44%.
Note: This calculation assumes that every refrigerator is equally likely to be chosen and that the mix-up on the production line occurred randomly.
To find the probability of at least 2 of the 5 selected refrigerators being defective, we can use the concept of combinations and the probability of selecting defective and non-defective refrigerators.
Step 1: Calculate the total number of ways to choose 5 refrigerators out of the total 50 (6 defective + 44 good) refrigerators. This can be calculated using the combination formula:
C(n, r) = n! / (r!(n-r)!), where n is the total number of items, and r is the number of items chosen.
In this case, n = 50 (total refrigerators) and r = 5 (selected refrigerators).
C(50, 5) = 50! / (5!(50-5)!) = 2,118,760
Step 2: Calculate the number of ways to choose exactly 0 defective refrigerators and 5 good ones, and 1 defective refrigerator and 4 good ones.
- For exactly 0 defective refrigerators and 5 good ones, we choose all 5 refrigerators from the good ones. The number of ways is C(44, 5) = 44! / (5!(44-5)!) = 1,086,008.
- For exactly 1 defective refrigerator and 4 good ones, we choose 1 defective refrigerator out of the 6 defective ones and 4 good refrigerators out of the 44 good ones. The number of ways is C(6, 1) * C(44, 4) = 6! / (1!(6-1)!) * 44! / (4!(44-4)!) = 6 * 44! / (5! * 4! * 40!) = 7,920.
Step 3: Calculate the probability of selecting at least 2 defective refrigerators.
Probability = (number of ways to choose exactly 0 defective refrigerators and 5 good ones + number of ways to choose exactly 1 defective refrigerator and 4 good ones) / total number of ways to choose 5 refrigerators.
Probability = (1,086,008 + 7,920) / 2,118,760
Probability ≈ 0.517 or 51.7%
Therefore, the probability that at least 2 out of the 5 selected refrigerators are defective is approximately 0.517 or 51.7%.