Explanation
$$n=5\\P(X=4)=^5C_4p^4q=5p^4q\\P(X=3)=^5C_3p^3q^2=10p^3q^2\\P(X=3)=P(X=4)\implies p=2q\\p+q=1\\q=\dfrac 13,p=\dfrac 23\\P(X>2)=P(X=3)+P(X=4)+P(X=5)\\10\left(\dfrac 23\right)^3\left(\dfrac 13\right)^2+5\left(\dfrac 23\right)^4\left(\dfrac 13\right)+\left(\dfrac 23\right)^5=\dfrac{192}{243}=0.79$$
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