Answer: 0.4083
Step-by-step explanation:
Let D be the event of receiving a defective television.
Given : The probability that the television is defective :-
[tex]P(D)=\dfrac{3}{13}[/tex]
The formula for binomial distribution :-
[tex]P(X=x)=^nC_xp^x(1-p)^{n-x}[/tex]
If two televisions are randomly selected, compute the probability that both televisions work, then the probability at least one of the two televisions does not work is given by :_
[tex]P(X\geq1)=P(1)+P(2)\\\\=^2C_1(\frac{3}{13})^1(1-\frac{3}{13})^{2-1}+^2C_2(\frac{3}{13})^2(1-\frac{3}{13})^{2-2}\\\\=0.408284023669\approx0.4083[/tex]
Hence , the required probability = 0.4083
