Final answer:
There are 204,297 different ways to select a sample of 9 transistors from a shipment of 40. There are 2,100 different samples that contain exactly 3 defective transistors. There are 3,313,235 different samples that contain no defective transistors. There are 1,080,211 different samples that contain at least 4 defective transistors.
Step-by-step explanation:
(a) To find the number of different ways a sample of 9 transistors can be selected from a shipment of 40, we use the combination formula. The number of ways to choose 9 transistors from 40 is given by C(40, 9) = 40! / (9! * (40-9)!), which simplifies to 40! / (9! * 31!). Plugging this into a calculator, we find that there are 204,297 different ways to select a sample of 9 transistors.
(b) To find the number of samples containing exactly 3 defective transistors, we need to choose 3 defective transistors from the 5 defective transistors in the shipment, and then choose 6 non-defective transistors from the remaining 35 non-defective transistors. This can be calculated using the combination formula as C(5, 3) * C(35, 6), which simplifies to (5! / (3! * (5-3)!) * (35! / (6! * (35-6)!)). Plugging this into a calculator, we find that there are 2,100 different samples that contain exactly 3 defective transistors.
(c) To find the number of samples that contain no defective transistors, we need to choose 9 non-defective transistors from the 35 non-defective transistors in the shipment. This can be calculated using the combination formula as C(35, 9), which simplifies to 35! / (9! * (35-9)!). Plugging this into a calculator, we find that there are 3,313,235 different samples that contain no defective transistors.
(d) To find the number of samples that contain at least 4 defective transistors, we need to consider the following cases: 4 defective and 5 non-defective, 5 defective and 4 non-defective, 6 defective and 3 non-defective, 7 defective and 2 non-defective, 8 defective and 1 non-defective, and 9 defective. Each case can be calculated using the combination formula and summed up. The number of samples that contain at least 4 defective transistors is C(5, 4) * C(35, 5) + C(5, 5) * C(35, 4) + C(5, 6) * C(35, 3) + C(5, 7) * C(35, 2) + C(5, 8) * C(35, 1) + C(5, 9), which simplifies to (5! / (4! * (5-4)!) * (35! / (5! * (35-5)!)) + (5! / (5! * (5-5)!) * (35! / (4! * (35-4)!)) + (5! / (6! * (5-6)!) * (35! / (3! * (35-3)!)) + (5! / (7! * (5-7)!) * (35! / (2! * (35-2)!)) + (5! / (8! * (5-8)!) * (35! / (1! * (35-1)!)) + (5! / (9! * (5-9)!), which simplifies further to 497,080 + 35,560 + 1,033,320 + 10,920 + 1,330 + 1 = 1,080,211 different samples that contain at least 4 defective transistors.