Question

Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randomly​ selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not​ work?

Answer 1

Answer:

a)

The probability that both televisions work is:  0.42

b)

The probability at least one of the two televisions does not​ work is:

                          0.5833

Step-by-step explanation:

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

          9-3=6

a)

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

b)

The probability at least one of the two televisions does not​ work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$

and the probability both do not work is:

$=\dfrac{3_C_2}{9_C_2}\\\\\\=\dfrac{1}{12}\\\\\\=0.0833$

Hence, Probability that atleast does not work is:

             0.5+0.0833=0.5833