A bin of 35 manufactured parts contains 6 defective parts and 29 non-defective
parts. A sample of size 5 parts is selected from 35 parts. Selected parts are not replaced. How
many different samples are there of size five that contain exactly 2 defective parts? What is
the probability that a sample contains exactly 2 defective parts?
To find out how many different samples of size five contain exactly two defective parts, we can use the combination formula.
The combination formula is given by:
C(n, r) = n! / (r! * (n-r)!)
Where n is the total number of parts, r is the number of defective parts, and C(n, r) denotes the number of ways to choose r items from a set of n items without regard to their order.
In this scenario, n = 35 (total number of parts) and r = 2 (number of defective parts). Plugging these values into the combination formula, we have:
C(35, 2) = 35! / (2! * (35-2)!)
= (35 * 34 * 33!) / (2! * 33!)
= (35 * 34) / (2 * 1)
= 595
Therefore, there are 595 different samples of size five that contain exactly two defective parts.
To find the probability that a sample contains exactly two defective parts, we need to calculate the ratio of the number of samples with exactly two defective parts to the total number of samples.
The total number of samples of size five can be calculated using the combination formula again, this time with r = 5 (since we want size five samples):
C(35, 5) = 35! / (5! * (35-5)!)
= (35 * 34 * 33 * 32 * 31!) / (5! * 30!)
= (35 * 34 * 33 * 32) / (5 * 4 * 3 * 2 * 1)
= 35 * 34 * 33 * 32
= 1,256,640
So, the probability of a sample containing exactly two defective parts is:
P(exactly 2 defectives) = C(35, 2) / C(35, 5)
= 595 / 1,256,640
= 0.0004732 (rounded to 6 decimal places)
≈ 0.04732% (rounded to 2 decimal places)
To determine the number of different samples of size five that contain exactly two defective parts, we can use the concept of combinations.
First, let's calculate the number of ways we can select two defective parts from the six defective parts. This can be done using the combination formula, which is denoted as C(n, r), where n is the total number of items and r is the number of items to be chosen. In this case, n = 6 and r = 2. So, the calculation for selecting two defective parts from six would be:
C(6, 2) = 6! / (2! * (6 - 2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15
Next, let's calculate the number of ways we can select three non-defective parts from the twenty-nine non-defective parts. This can be calculated using the combination formula as well, with n = 29 and r = 3. The calculation would be:
C(29, 3) = 29! / (3! * (29 - 3)!) = 29! / (3! * 26!) = (29 * 28 * 27) / (3 * 2 * 1) = 3654
Since we are selecting one sample, we need to multiply the two calculated values together:
Number of samples = C(6, 2) * C(29, 3) = 15 * 3654 = 54760
Therefore, there are 54,760 different samples of size five that contain exactly two defective parts.
To calculate the probability that a sample contains exactly two defective parts, we need to divide the number of samples that contain exactly two defective parts by the total number of possible samples.
Total possible samples can be calculated using the combination formula with n = 35 and r = 5:
C(35, 5) = 35! / (5! * (35 - 5)!) = 35! / (5! * 30!) = (35 * 34 * 33 * 32 * 31) / (5 * 4 * 3 * 2 * 1) = 324632
So, the probability can be calculated as:
Probability = Number of samples with exactly two defective parts / Total number of possible samples
= 54,760 / 324,632
≈ 0.169 or 16.9% (rounded to two decimal places)
Therefore, the probability that a sample contains exactly two defective parts is approximately 0.169 or 16.9%.