From a shipment of 75 transistors, 4 of which are defective, a sample of 6 transistors is selected at random.
(b) How many samples contain exactly 3 defective transistors?
(c) How many samples do not contain any defective transistors?
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What piece of info is missing?
To answer these questions, we can use the concept of combinations.
(b) To find the number of samples that contain exactly 3 defective transistors, we need to choose 3 defective transistors from the 4 defective ones and 3 non-defective transistors from the remaining 71 non-defective ones. The number of ways to do this is given by the formula for combinations:
C(n, r) = n! / (r! * (n-r)!)
In this case, n represents the total number of items (75) and r represents the number of items we want to choose (6). Therefore, the number of samples that contain exactly 3 defective transistors can be calculated as:
C(4, 3) * C(71, 3) = (4! / (3! * (4-3)!)) * (71! / (3! * (71-3)!))
Simplifying this expression will give us the answer.
(c) To find the number of samples that do not contain any defective transistors, we need to choose 6 non-defective transistors from the 71 available non-defective ones. Again, this can be calculated using the formula for combinations:
C(n, r) = n! / (r! * (n-r)!)
In this case, n represents the total number of items (71) and r represents the number of items we want to choose (6). Therefore, the number of samples that do not contain any defective transistors can be calculated as:
C(71, 6) = (71! / (6! * (71-6)!))
Simplifying this expression will give us the answer.