Two two-digit counting numbers (from 10 to 99, inclusive) were chosen randomly.These two integers could be the same or different. What is the probability that thedifference between these two numbers is at most 3?
+9519 There are (99 - 10 + 1)^2 = 8100 ways to choose two such numbers.
99 - 10 + 1 = 90 cases out of all cases has two same numbers chosen,
2(98 - 10 + 1) = 178 cases out of all cases has two numbers with difference = 1.
2(97 - 10 + 1) = 176 cases out of all cases has two numbers with difference = 2.
2(96 - 10 + 1) = 174 cases out of all cases has two numbers with difference = 3.
Therefore, the required probability is \(\dfrac{90 + 178 + 176 + 174}{8100} = \dfrac{103}{1350}\).
