<s>[INST] 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.
 [/INST] 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 $\frac {149} {7^4} = \frac {149} {2401}.$
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