Suppose you just received a shipment of thirteen televisions. Two 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.)

Step 1 ：We have a total of 13 televisions, out of which 2 are defective and 11 are working.

Step 2 ：We are to select 2 televisions randomly and find the probability that both are working.

Step 3 ：The total number of ways to select 2 televisions from 13 is given by the combination formula \(C(n, r) = \frac{n!}{r!(n-r)!}\) where n is the total number of items, r is the number of items to choose, and '!' denotes factorial. In this case, n=13 and r=2. This gives us a total of 78 ways.

Step 4 ：The number of ways to select 2 working televisions from the 11 that are working is also given by the combination formula. In this case, n=11 and r=2. This gives us a total of 55 ways.

Step 5 ：The probability that both televisions work is then given by the ratio of these two quantities, which is \(\frac{55}{78} = 0.705\).

Step 6 ：Final Answer: The probability that both televisions work is \(\boxed{0.705}\).