Question on drawing 4 tickets form 7 tickets A bag contains 7 tickets marked with the numbers 0, 1, 2, ..., 6 respectively. A ticket is drawn and replaced; find the chance that after 4 drawings the sum of the numbers drawn is 8.
 A: The tickets are drawn after replacement. Hence, the total number of possible ways of draws is 7 x 7x 7x 7 = 2401.
Now, for the number of ways in which the total adds up to 8 is the subset of these 2401 possible ways where the total is 8. For example, {3, 4, 0, 1} or {3, 2, 2, 1} adds to 8. 
The total number of ways can either be counted by brute-force or recursively. Recursively, the number of ways to get a sum S with k drawings is: Sum over, each card value i from 0 to 6, number of ways for sum $(S-i)$ with $(k-1)$ drawings
$$ Count(S, k) = \sum_{i=0}^{6} Count(S-i, k-1) $$
The number of such draws is 149. 
$$ P(Sum = 8) = 149/2401 $$
A: We can view $4$ drawings to be $4$ different boxes and the number on the ticket drawn at $i$-th drawing is the number of balls in the $i$-th box after the balls are thrown to the boxes where $1 \leq i \leq 4.$ Now since the numbers on the tickets are from $0,1, 2 , \cdots ,6$ so that means each box will contain at most $6$ balls.
So the number of ways to get $8$ as the sum of the tickets marked $0,1,2,.\cdots , 6$ in $4$ drawings is the number of ways $n$ to put $8$ indistinguishable balls in $4$ different boxes so that each box will contain at most $6$ balls.
Now let $n_7$ denote the number of ways to put $8$ indistinguishable balls in $4$ different boxes in which one of the boxes will contain $7$ balls and $n_8$ denote the number of ways to put $8$ indistinguishable balls in $4$ different boxes in which one of the boxes will contain $8$ balls. Then clearly $n_7 = 12$ and $n_8 = 4.$ 
Let $m$ be the number of ways to put $8$ indistinguishable balls in $4$ different boxes. Then $m = \binom {11} {3} = 165.$
Then $n = m - n_7 - n_8 = 165 - 12 - 4 = 149.$
So the required probability is $\frac {149} {7^4} = \frac {149} {2401}.$
