Suppose you just received a shipment of eleven 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:  The probability that both televisions work : 0.5329The  probability at least one of the two televisions does not​ work : 0.4671Step-by-step explanation:Given : The total number of television : 11The number of defective television : 3The probability that the television is defective : [tex]p=\dfrac{3}{11}\approx0.27[/tex] Binomial distribution formula :- [tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where P(X) is the probability of getting success in x trials, p is the probability of success and n is the total trials.If two televisions are randomly​ selected, then the probability that both televisions work:[tex]P(0)=^2C_0(0.27)^0(1-0.27)^{2-0}=(1)(0.73)^2=0.5329[/tex]The probability at least one of the two televisions does not​ work :[tex]P(X\geq1)=1-P(0)=1-0.5329=0.4671[/tex]