7. A die is rolled twice in succession, and the numbers obtained are $a$ and $b$ respectively. Then the probability $p=$ $\qquad$ that the cubic equation $x^{3}-(3 a+1) x^{2}+(3 a+2 b) x-2 b=0$ has three distinct real roots is $\qquad$ .(Answer with a number).
Solve
$x^{3}-(3 a+1) x^{2}+(3 a+2 b) x-2 b=(x-1)\left(x^{2}-3 a x+2 b\right)$, then $9 a^{2}-8 b>0$ and $3 a \neq 2 b+1$. Therefore, when $b=1$, $a=2,3,4,5,6 ; b=2$  when, $a=2,3,4,5,6 ; b=3$ when, $a=2,3,4,5,6 ; b=4$  when, $a=2,4,5,6 ; b=5$  when, $a=3,4,5,6 ; b=6$  when, $a=3,4,5,6$. Let $A=$ “the equation has three distinct real roots”, so $P(A)=\frac{27}{36}=\frac{3}{4}$.