Suppose the 5 friends are A,B,C,D,E, where D is Dhruv.

In the "successful" cases, some of them will not get any chocolates, if Dhruv
has to get at least 6 and there are only 8 chocolates.

There are 4 cases where Dhruv gets at least 6 chocolates:

Case 1.  Dhruv gets 6 chocolates and 1 of the other 4 gets both the other 2
chocolates. Choose the 1 friend to get the other 2 chocolates in 4 ways.

Case 2.  Dhruv gets 6 chocolates and 2 of the other 4 get 1 chocolate each.
Choose the 2 friends to get 1 chocolate each in C(4,2) = 6 ways.

Case 3.  Dhruv gets 7 chocolates and 1 of the other 4 gets the other 1 chocolate.
Choose the 1 friend to get the other 1 chocolate in 4 ways.  

Case 4.  Dhruv gets all 8 chocolates and the other 4 get none.  That's 1 way.

That's 4+6+4+1=15 "successful" cases.

To find all the possibilities, make a line of 8 stars and 4 bars, like this
sample illustration:

***||***|**|

A gets the same number of chocolates as the number of bars left of the left-most
bar, which in the sample illustration is 3
B gets the same number of chocolates as the number of stars between the 1st and
2nd bars, which in the sample illustration is 0.
C gets the same number of chocolates as the number of stars between the 2nd and
3rd bars, which in the sample illustration is 3.
D gets the same number of chocolates as the number of stars between the 3rd and
4th bars, which in the sample illustration is 2.
E gets the same number of chocolates as the number of stars to the right of the
4th bar, which in the sample illustration is 0.

There are 8 stars and 4 bars.  That's 12 things, where the 8 stars are
indistinguishable from each other, and also the 4 bars are also indistinguishable from each other.

So the number of ways the chocolates can be divided is .

So the probability that Dhruv gets at least 6 chocolates is 15/495 which reduces
to 1/33.

Edwin