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A shipment of 105 laptops contains 5 defective laptops. A quality control specialist chooses a sample of 10 laptops from the shipment.
(a) How many possible choices of 10 laptops can be made?
(b) How many of these possible selections will not contain any defective laptops?
(c) How many of the possible selections will contain at least one defective laptop?
(d) How many of the possible selections will contain exactly one defective laptop?
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In the post by @MathLover1, the solutions, the numbers and the answers
to (b), (c) and (d) are incorrect.
I came to bring a correct solution.
(a) How many possible choices of 10 laptops can be made?
C(105,10) = 28848458598960.
(b) How many of these possible selections will not contain any defective laptops?
C((105-5),10) = C(100,10) = 17310309456440.
It is the number of all possible combinations of 10 laptops from 105-5 = 100 good laptops.
(c) How many of the possible selections will contain at least one defective laptop?
C(105,10) - C(100,10) = 28848458598960 - 17310309456440 = 11538149142520.
From the total number of all possible combinations of 10 laptops from 105 laptops,
we subtract the number of all possible combinations of 10 laptops from 105-5 = 100 good laptops.
(d) How many of the possible selections will contain exactly one defective laptop?
C(5,1)*C(100,9) = 5*1902231808400 = 9511159042000.
In this assignment, we combine 1 defective laptop from 5 defectives laptops
with 9 good laptops from 105-5 = 100 good laptops.
Solved (correctly).