Question

A shipment of 25 television sets contains three defective units. In how many ways can a vending company purchase four of these units and receive (a) all good units, (b) two good units, and (c) at least two good units?

Answer 1

Answer:

a) 1540 ways

b) 693 ways

c) 2233 ways

Step-by-step explanation:

The order is not important. For example, buying TV LG A, LG B and LG C is the same as buing LG B, then LG A, then LG C.

So we use the combinations formula to solve this problem:

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

A shipment of 25 television sets contains three defective units.

This means that three are defective and 22 are ok.

In how many ways can a vending company purchase four of these units and receive

(a) all good units

This is a selection of 3 from a set of 22(all ok).

So

$C_{22,3} = \frac{22!}{3!*19!} = 1540$

(b) two good units

This is 3(number of defectives) multiplied by a selection of 2 from a set of 22(all ok).

So

$3*C_{22,2} = 3*\frac{22!}{2!*20!} = 693$

(c) at least two good units?

This is the sum of a and b, since there are three units, and at least two is two or three.

So

1540 + 693 = 2233