A box contains coupons labeled 1,2,3 ....n. A coupon is picked at random and the number x is noted. The coupon is put back into the box and a new coupon is picked at random. The new number is y . Then the probability that one of the numbers x , y divides the other is (in the options below [r] denotes the largest integer less than or equal to r )

Step-by-step solution

Let x=1 favourable out comes (1,1),(1,2) (1, n) no. of favourable out comes when x=1 = [ n 1 ] no. of favourable out comes when x=1 or y=1 =2 [ n 1 ]-1 no. of favourable out comes when x=2 or y=2 but x 1, y 1 2 [ n 2 ]-1 Similarly no. of favourable out comes when x=k or y=k but x, y 1,2, . . k-1 2 [ n k ]-1 So probability = _ k=1 ^ n [ n k ]-(1+1 . . n times ) n² = 2 n ² _ k =1 ^ n [ n k ]- 1 n