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A special dice with numbers $1,-1,2,-2,0$ and 3 is thrown thrice. What is the probability that the sum of the numbers occurring on the upper face is zero?
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$25 / 216$
Total no. of elementary events $=6^{3}$. Favourable no. of elementary events
$=$ coefficient of $x^{0}$ in $\left[x+x^{-1}+x^{0}+x^{-2}+x^{2}+x^{3}\right]^{3}$
$=$ coeff. of $x^{0}$ in $\left[\frac{1+x+x^{2}+x^{3}+x^{4}+x^{5}}{x^{2}}\right]^{3}$
= coeff. of $x^{6}$ in $\left[1+x+x^{2}+x^{3}+x^{4}+x^{5}\right]^{3}$
$=$ coeff. of $x^{6}$ in $\left[\frac{1-x^{6}}{1-x}\right]^{3}$
$=$ coeff. of $x^{6}$ in $\left[1-x^{6}\right]^{3}[1-x]$
$=$ coeff. of $x^{6}$ in $\left[1-{ }^{3} C_{1} x^{6}+\ldots .\right][1-x]^{-3}$
$=$ coeff. of $x^{6}$ in $(1-x)^{-3} \cdot{ }^{3} C_{1}$ coeff. of $x^{0}$ in $(1-x)^{-3}$
$={ }^{6+3-1} C_{3-1}-{ }^{3} C_{1}$
$={ }^{8} C_{2}-{ }^{3} C_{1}=\frac{8 !}{6 ! 2 !}-\frac{3 !}{2 !}$
$=\frac{8 \times 7}{2}-3=25$
Required probability $=\frac{25}{216}$
$=$ coefficient of $x^{0}$ in $\left[x+x^{-1}+x^{0}+x^{-2}+x^{2}+x^{3}\right]^{3}$
$=$ coeff. of $x^{0}$ in $\left[\frac{1+x+x^{2}+x^{3}+x^{4}+x^{5}}{x^{2}}\right]^{3}$
= coeff. of $x^{6}$ in $\left[1+x+x^{2}+x^{3}+x^{4}+x^{5}\right]^{3}$
$=$ coeff. of $x^{6}$ in $\left[\frac{1-x^{6}}{1-x}\right]^{3}$
$=$ coeff. of $x^{6}$ in $\left[1-x^{6}\right]^{3}[1-x]$
$=$ coeff. of $x^{6}$ in $\left[1-{ }^{3} C_{1} x^{6}+\ldots .\right][1-x]^{-3}$
$=$ coeff. of $x^{6}$ in $(1-x)^{-3} \cdot{ }^{3} C_{1}$ coeff. of $x^{0}$ in $(1-x)^{-3}$
$={ }^{6+3-1} C_{3-1}-{ }^{3} C_{1}$
$={ }^{8} C_{2}-{ }^{3} C_{1}=\frac{8 !}{6 ! 2 !}-\frac{3 !}{2 !}$
$=\frac{8 \times 7}{2}-3=25$
Required probability $=\frac{25}{216}$
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