Question

suppose you just received a shipment of eleven televisions. four 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? the probability that both televisions work is (round to three decimal places as needed.)

Explanation:

Step1: Calculate the number of working televisions

Total televisions \(n = 11\), defective televisions \(d=4\). So working televisions \(w=11 - 4=7\).

Step2: Calculate the probability that the first television works

The probability that the first - selected television works is \(P_1=\frac{7}{11}\).

Step3: Calculate the probability that the second television works (given the first works)

After one working television is selected, there are \(n_1 = 10\) televisions left and \(w_1=6\) working televisions left. So the probability that the second - selected television works given the first works is \(P_2=\frac{6}{10}\).

Step4: Calculate the probability that both work

By the multiplication rule for dependent events \(P = P_1\times P_2=\frac{7}{11}\times\frac{6}{10}=\frac{42}{110}=\frac{21}{55}\approx0.382\)

Answer:

\(0.382\)