Three numbers are chosen from 1 to 30. The probability that minimum is 10 and maximum is 26 is _______

Step-by-step solution

Total number of ways to select 3 numbers from 30 is ³⁰C₃ . ³⁰C₃ = 30 29 28 3 2 1 = 4060 For the minimum to be 10 and the maximum to be 26, two of the selected numbers must be exactly 10 and 26. The third number must be chosen from the integers strictly between 10 and 26. The integers between 10 and 26 are 11, 12, ..., 25. The number of such integers is 26 - 10 - 1 = 15 . Thus, the number of favorable selections is 15. The required probability is 15 4060 = 3 812 . Answer: 3 812