Question

Counters marked $1,2,3$ are placed in a bag, and one is withdrawn and replaced. The operation being repeated three times, what is the chance of obtaining a total of $6$?

Answer 1

Possible number of outcomes $=3^3=27$

Chance of obtaining a total $6$ $=$ coefficient of $x^6$ in ${(x+x^2+x^3)}^3$

${(x+x^2+x^3)}^3=x^3{(1+x+x^2)}^3=x^3{\left(\frac{x^3-1}{x-1}\right)}^3=x^3{(x^3-1)}^3{(x-1)}^{-3}$

Now chance of obtaining $6$ $=$ coefficeint of $x^3$ in ${(x^3-1)}^3{(x-1)}^{-3}$

${(x^3-1)}^3{(x-1)}^{-3}={(x}^9-1-3x^6+3x^3\left)\right(1+3x-6x^2+10x^3....)$

$\therefore$ coefficient of $x^3=10+9=19$

$\therefore$ desired probability $=\frac{19}{27}$