If three dice are thrown simultaneously, then the probability of getting a score of 7 is

$\frac{5}{72}$

$n(S)= 6 ×6×6$

n(E) = The number of solutions of x + y + z = 7 ,

where $1 ≤ x ≤ 5, 1 ≤ y ≤ 5, 1 ≤ z ≤ 5 $

= Coefficient of $x^7$ in $(x + x^2 + ... +x^5)^3$

= Coefficient of $x^4$ in $(1+x + ... + x^4)^3 =$ Coefficient of $x^4$ in $\left(\frac{1-x^5}{1-x}\right)^3$

= Coefficient of $x^4$ in $(1- 3x^5 + 3x^{10} - x^{15}) (1-x)^{-3}$

= Coefficient of $x^4$ in $(1- 3x^5 + 3x^{10} - x^{15})({^2C}_0 + {^3C}_1x + {^4C}_2x^2 +  {^5C}_3x^3+{^6C}_4 x^4 + ...)$

$= {^6C}_4 = \frac{6!}{4!2!}=\frac{6×5}{2}=15.$

$∴ P(E) = \frac{n(E)}{n(S)}=\frac{15}{6×6×6}=\frac{5}{72}.$