Suppose you just received a shipment of thirteen 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?
Accepted Solution
A:
Answer: 0.4083Step-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