A bag contains 10 balls numbered from 0 to 9. the balls are such that the person picking a ball out of the bag is equally likely to pick anyone of them. A person picked a ball and replaced it in the bag after noting its number. He repeated this process 2 more times. What is the probability that the ball picked first is numbered higher than the ball picked second and the ball picked second is numbered higher than the ball picked third?  Choose the correct choice from the following answers:	A) 72/100	B) 3/25	C) 4/5	D) 1/6	E) 1/9
Explanation :   Let, the number of the ball picked first =a, second =b and third =c and these three numbers a, b and c are distinct.   Three distinct ball can be picked in (10×9×8) ways.   The order of a, b and c can be as follows:-   (i) a&gt;b&gt;c.   (ii) a&gt;c&gt;b.   (iii) b&gt;c&gt;a.   (iv) b&gt;a&gt;c.   (v) c&gt;a&gt;b.   (vi) c&gt;b&gt;a.   They will occur equal number of times. So, the number of ways in which (a&gt;b&gt;c) are :-   =&gt;1/6 x 10 x 9 x 8.   =&gt;120.   Hence, the required probability is :-   =&gt;120 / (10x10x10).   =&gt;3/25.   Answer : 3/25
<result>B</result>
A bag contains 10 balls numbered from 0 to 9. the balls are such that the person picking a ball out of the bag is equally likely to pick anyone of them. A person picked a ball and replaced it in the bag after noting its number. He repeated this process 2 more times. What is the probability that the ball picked first is numbered higher than the ball picked second and the ball picked second is numbered higher than the ball picked third?