Question

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 $8$.

Answer 1

Draw 1	Draw 2	Draw 3	Draw 4	Number of Ways
6	2	0	0	$\frac{4!}{2!}=12$
6	1	1	0	$\frac{4!}{2!}=12$
5	3	0	0	$\frac{4!}{2!}=12$
5	2	1	1	$\frac{4!}{2!}=12$
5	1	1	1	$\frac{4!}{3!}=4$
4	4	0	0	$\frac{4!}{2!2!}=6$
4	3	1	0	$4!=24$
4	2	2	0	$\frac{4!}{2!}=12$
4	2	1	1	$\frac{4!}{2!}=12$
3	3	2	0	$\frac{4!}{2!}=12$
3	3	1	1	$\frac{4!}{2!2!}=6$
3	2	2	1	$\frac{4!}{2!}=12$
2	2	2	2	$\frac{4!}{4!}=1$

$\therefore$Number of ways for the sum to be 8 after 4 draws $=149$

Total Number of ways of 4 draws $=7^4$

$\therefore$ Probability of sum of numbers to be 8 after 4 drawings $=\frac{149}{7^4}$