Question

Two integers x and y are chosen with
replacement from the set {0, 1, 2, ..., 10). The
probability that |x-y/ > 5 is
mode of x-y greater than 5
​

Answer 1

Answer with explanation:

Number of elements in the set = {0, 1, 2, ..., 10}

Number of ordered pairs(a,b) ,from the elements of set ,such that , |a-b|>5,

={(0,10),(0,9),(0,8),(0,7),(0,6),(1,10),(1,9),(1,8),(1,7),(2,10),(2,9),(2,8),(3,10),(3,9),(4,10),(10,0),(9,0),(8,0),(7,0),(6,0),(10,1),(9,1),(8,1),(7,1),(10,2),(9,2),(8,2),(10,3),(9,3),(10,4)}="30" in number

Total number of ordered pair,if there are 11 number in all , from, 0 to 11,as order of arrangement of numbers is important,

   $=_{2}^{11}\textrm{P}=\frac{11!}{9!}=\frac{11\times 10\times 9!}{9!}=110$

Required Probability,means probability that |x-y| > 5 ,or mode of x-y, is greater than 5

 $=\frac{\text{Total favorable outcome}}{\text{Total possible outcome}}\\\\=\frac{30}{110}\\\\=\frac{3}{11}$