Respuesta :
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:
[tex]=\dfrac{6_C_2}{9_C_2}[/tex]
( 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:
[tex]=\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[/tex]
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:
[tex]=\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[/tex]
and the probability both do not work is:
[tex]=\dfrac{3_C_2}{9_C_2}\\\\\\=\dfrac{1}{12}\\\\\\=0.0833[/tex]
Hence, Probability that atleast does not work is:
0.5+0.0833=0.5833
