A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is $\frac{k}{3^{11}}$, then $k$ is equal to :

Given, a pair of dice is thrown once, then number of outcomes $=6 \times 6=36$ <br/><br/>A total of 5 are $\{(1,4),(2,3),(4,1),(3,2)\}$ <br/><br/>$$ \begin{aligned} & \therefore p =p(\text { success })=\frac{4}{36}=\frac{1}{9} \\\\ & q =1-\frac{1}{9}=\frac{8}{9} \end{aligned} $$ <br/><br/>Let $X$ represent the number of success <br/><br/>$$ \begin{aligned} & \therefore P(\text { at least } 4 \text { success }) \\\\ &=P(X=4)+P(X=5) \\\\ &={ }^5 C_4 p^4 q+{ }^5 C_5 p^5 q^{5-5} \\\\ &=5\left(\frac{1}{9}\right)^4\left(\frac{8}{9}\right)+\left(\frac{1}{9}\right)^5 \\\\ &=\frac{40+1}{\left(3^2\right)^5}=\frac{41}{310} \\\\ &=\frac{41 \times 3}{3^{11}}=\frac{123}{3^{11}} \\\\ & \therefore k=123 \end{aligned} $$