Question

A shipment of 150 television sets contains ten defective units. In how many ways can a vending company buy five of theses units and receive a. No defective units b. One defective unit. c. At least one defective unit.

Answer 1

The number of ways a vending company can buy five television units with no defective units is 2,598,960. For one defective unit, it's 10 * C(140,4), and for at least one defective unit, it's the total ways to choose five sets minus the ways for no defective units.

Step-by-step explanation:

The question from the student involves combinatorics, a branch of mathematics dealing with counting, both as a means and an end in obtaining results, and certain properties of finite structures. Specifically, the student is asking about the ways to select television units without or with defective ones, which is an example of a hypergeometric distribution problem.

To calculate the various ways to select television sets:

1. No defective units: We use the combination formula to choose 5 from the 140 non-defective units (since the total is 150 and there are 10 defective, 150 - 10 = 140).
   C(140, 5) = 140! / (5! * (140 - 5)!) = 2,598,960 ways.
2. One defective unit: We choose one from the 10 defective units and four from the 140 non-defective units.
   C(10,1) * C(140,4) = 10 * (140! / (4! * (140 - 4)!)) = 10 * 4,845,354 ways.
3. At least one defective unit: This can be found by subtracting the number of ways to select sets with no defective units from the total ways to choose any five sets.
   C(150, 5) - C(140, 5) = (150! / (5! * (150 - 5)!)) - 2,598,960 ways.

Note that '!' denotes factorial, which is the product of all positive integers up to that number.