From a shipment of 75 transistors, 6 of which are defective, a sample of 5 transistors is selected at random.
(a) In how many different ways can the sample be selected?
ways
(b) How many samples contain exactly 3 defective transistors?
samples
(c) How many samples do not contain any defective transistors?
samples
(a) 75C5
(b) 69C2 * 6C3
(c) 69C5
In B and C where did the 69 come from?
in B where did the also come from
(a) In how many different ways can the sample be selected?
To find the number of ways the sample of 5 transistors can be selected from a shipment of 75 transistors, we can use the combination formula:
C(n, r) = n! / (r! * (n-r)!)
Where:
n = total number of items in the shipment (75)
r = number of items selected for the sample (5)
C(75, 5) = 75! / (5! * (75-5)!)
= 75! / (5! * 70!)
= (75 * 74 * 73 * 72 * 71) / (5 * 4 * 3 * 2 * 1)
= 216,217,024
So, there are 216,217,024 different ways the sample can be selected.
(b) How many samples contain exactly 3 defective transistors?
Since there are 6 defective transistors in the shipment, we need to choose 3 of them for the sample. The remaining 2 transistors should be non-defective. We can calculate this using the combination formula:
C(6, 3) = 6! / (3! * (6-3)!)
= 6! / (3! * 3!)
= 6 * 5 * 4 / (3 * 2 * 1)
= 20
Now, we need to choose the remaining 2 non-defective transistors from the remaining 75-6 = 69 transistors. Again, we can use the combination formula:
C(69, 2) = 69! / (2! * (69-2)!)
= 69! / (2! * 67!)
= 69 * 68 / (2 * 1)
= 2,346
Finally, we can find the total number of samples containing exactly 3 defective transistors by multiplying the two calculations:
Total samples = C(6, 3) * C(69, 2)
= 20 * 2,346
= 46,920
So, there are 46,920 samples that contain exactly 3 defective transistors.
(c) How many samples do not contain any defective transistors?
Since there are 6 defective transistors in the shipment, we need to choose all 5 non-defective transistors for the sample. We can calculate this using the combination formula:
C(75-6, 5) = C(69, 5)
= 69! / (5! * (69-5)!)
= 69! / (5! * 64!)
= 69 * 68 * 67 * 66 * 65 / (5 * 4 * 3 * 2 * 1)
= 13,537,080
So, there are 13,537,080 samples that do not contain any defective transistors.
To answer these questions, we need to understand the concept of combinations and apply the formula for combinations.
(a) To calculate the number of different ways to select a sample of 5 transistors out of 75, we can use the formula for combinations, which is:
C(n, r) = n! / (r! * (n - r)!)
In this case, n = 75 (total number of transistors) and r = 5 (number of transistors to be selected in the sample).
Using the formula, we can calculate the number of different ways to select the sample:
C(75, 5) = 75! / (5! * (75 - 5)!)
= 75! / (5! * 70!)
= (75 * 74 * 73 * 72 * 71) / (5 * 4 * 3 * 2 * 1)
= 3,345,184 different ways
Therefore, there are 3,345,184 different ways to select the sample of 5 transistors out of 75.
(b) To calculate the number of samples that contain exactly 3 defective transistors, we can again use the formula for combinations.
We have 6 defective transistors out of 75, and we want to select exactly 3 of them in the sample.
Using the formula, we can calculate the number of different ways to select 3 defective transistors from 6:
C(6, 3) = 6! / (3! * (6 - 3)!)
= 6! / (3! * 3!)
= (6 * 5 * 4) / (3 * 2 * 1)
= 20 different ways to select 3 defective transistors.
Then, for the remaining 2 transistors, we select them from the non-defective transistors, which gives us:
C(75 - 6, 2) = 69! / (2! * (69 - 2)!)
= 69! / (2! * 67!)
= (69 * 68) / (2 * 1)
= 2,362 different ways to select 2 non-defective transistors from the remaining.
To find the total number of samples that contain exactly 3 defective transistors, we multiply these two values:
Total number of samples = 20 * 2,362 = 47,240
Therefore, there are 47,240 samples that contain exactly 3 defective transistors.
(c) To calculate the number of samples that do not contain any defective transistors, we need to select 5 transistors from the non-defective ones.
Since there are 75 transistors in total and 6 are defective, 69 are non-defective.
Using the formula for combinations, we can calculate the number of different ways to select 5 transistors from the 69 non-defective ones:
C(69, 5) = 69! / (5! * (69 - 5)!)
= 69! / (5! * 64!)
= (69 * 68 * 67 * 66 * 65) / (5 * 4 * 3 * 2 * 1)
= 6,172,763 different ways to select 5 non-defective transistors.
Therefore, there are 6,172,763 samples that do not contain any defective transistors.