Let us throw one dice three times, and let the number that comes up on the first throw be $a$, on the second throw be $b$, and on the third throw be $c$. Using these $a , b$ and $c$, we consider the quadratic function $f ( x ) = a x ^ { 2 } + b x + c$. (1) The probability that $b = 4$ and that the quadratic equation $f ( x ) = 0$ has two different real solutions is $\frac { \mathbf { N } } { \mathbf { O } \mathbf { P Q } }$. (2) Let us find the probability that $f ( 10 ) > 453$. The number of the cases of $( a , b , c )$ such that $f ( 10 ) > 453$ is as follows: when $a = 4$ and $b = 5$, it is $\mathbf { R }$; when $a = 4$ and $b = 6$, it is $\mathbf{S}$; when $a = 5$, it is $\mathbf{TU}$; when $a = 6$, it is $\mathbf{VW}$. Hence, the probability that $f ( 10 ) > 453$ is $\frac { \mathbf { X } } { \mathbf { Y } }$.
Let us throw one dice three times, and let the number that comes up on the first throw be $a$, on the second throw be $b$, and on the third throw be $c$. Using these $a , b$ and $c$, we consider the quadratic function $f ( x ) = a x ^ { 2 } + b x + c$.
(1) The probability that $b = 4$ and that the quadratic equation $f ( x ) = 0$ has two different real solutions is $\frac { \mathbf { N } } { \mathbf { O } \mathbf { P Q } }$.
(2) Let us find the probability that $f ( 10 ) > 453$.
The number of the cases of $( a , b , c )$ such that $f ( 10 ) > 453$ is as follows:\\
when $a = 4$ and $b = 5$, it is $\mathbf { R }$;\\
when $a = 4$ and $b = 6$, it is $\mathbf{S}$;\\
when $a = 5$, it is $\mathbf{TU}$;\\
when $a = 6$, it is $\mathbf{VW}$.\\
Hence, the probability that $f ( 10 ) > 453$ is $\frac { \mathbf { X } } { \mathbf { Y } }$.