A new shipment of 9 VCRs, which will be offered at a very low price, are delivered, Unknown to the manager, 4 of the VCRs are defective, Determine the probability that at least 1 of the first three VCRs sold is defective.Did I do this right? 1-5C3/9C3 =88.09% or do I do 1-(5/9)^3 = 82.9%. Please help I don't understand why the different methods are yielding a different answer. They should be the same. Help please!
To properly answer this question, let's break it down step by step:
Method 1: 1 - (5C3 / 9C3) = 88.09%
This method uses combinations to calculate the probability. Let's explain it in more detail:
1. First, we calculate the number of ways to choose 3 VCRs from a total of 9, which is denoted as 9C3 or "9 choose 3" and equals 84.
2. Next, we calculate the number of ways to choose 3 VCRs from the 5 remaining non-defective VCRs, denoted as 5C3 or "5 choose 3" and equals 10.
3. Finally, we subtract the number of ways to choose 3 non-defective VCRs from the total number of ways to choose any 3 VCRs (9C3). This gives us the number of ways to have at least one defective VCR among the first three sold.
The formula you used, 1 - (5C3 / 9C3), represents the probability of having at least one defective VCR among the first three sold. This calculations yields a probability of 88.09%.
Method 2: 1 - (5/9)^3 = 82.9%
This method uses the probability of not having a defective VCR, denoted as (5/9), and calculates the probability of this event not happening for three consecutive VCRs.
The probability of not having a defective VCR is 5 out of 9 since four of the nine VCRs are defective.
The formula you used, 1 - (5/9)^3, represents the probability of having at least one defective VCR among the first three sold. This calculations yields a probability of 82.9%.
The reason the two methods yield different results is that they are slightly different interpretations of the problem:
Method 1 calculates the number of ways to choose 3 VCRs, ensuring at least one is defective. This method considers the total number of 3 VCR combinations, regardless of order.
Method 2 calculates the probability of having at least one defective VCR in three consecutive sales. This method considers the order of sales and assumes each sale is independent.
Both methods are valid; however, the results can differ due to the slight difference in interpretation. It's important to choose the appropriate method based on the specific scenario you are analyzing.