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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?









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Option(B) is correct

Let the number on the ball picked first $= a$, second $=b$ and third $=c$

The 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). $a > b > c$ ; 
(ii). $a > c > b$ ; 
(iii). $b >c >a$ ; 
(iv). $b > a > c$ ; 
(v). $c > a > b$ ; 
(vi). $c > b > a$

They will occur equal number of times.

Thus the number of ways in which ($a > b > c$):

\(=\dfrac{1}{6}\times 10\times 9\times 8\)


Required probability

\(=\dfrac{120}{10\times 10\times 10}\)


(2) Comment(s)


Why have $a$, $b$, $c$ got to be distinct?

Are they representing the favourable cases?

The favourable cases should have 8 for $a$ and 7 for $b$ and 6 for $c$.

The solution needs to be more detailed.


They have to be distinct to support the next claim of division by 1/6.

Assume a sample case of 0,1,2. It can be permuted in 6 ways out of which only 2,1,0 is the favorable case.

Take 10*9*8 to be all such numbers which are different and 1/6 of these numbers are all such numbers which keep the invariant, $$a>b>c$$