Probability that the equation so formed will have real roots

Three dice named A,B,C are thrown and the numbers shown on them are put for $a,b,c$ respectively in the quadratic equation $ax^2+bx+c=0$.What is the probability that the equation so formed will have real roots?


Since the roots are real.Therefore,$b^2\geq 4ac$

When $b=2$,then $a=1,c=1$;When $b=3$,then $a=2,c=1$;When $b=3$,then $a=1,c=1$
When $b=3$,then $a=1,c=2$;When $b=4$,then $a=2,c=2$;When $b=4$,then $a=1,c=4$
When $b=4$,then $a=4,c=1$;When $b=4$,then $a=1,c=3$;When $b=4$,then $a=3,c=1$
When $b=4$,then $a=1,c=2$;When $b=4$,then $a=2,c=1$;When $b=4$,then $a=1,c=1$
When $b=5$,then $a=2,c=3$;When $b=5$,then $a=3,c=2$;When $b=5$,then $a=5,c=1$
When $b=5$,then $a=1,c=5$;When $b=5$,then $a=2,c=2$;When $b=5$,then $a=3,c=1$
When $b=5$,then $a=1,c=3$;When $b=5$,then $a=2,c=1$;When $b=5$,then $a=1,c=2$
When $b=5$,then $a=1,c=1$;When $b=6$,then $a=3,c=3$;When $b=6$,then $a=2,c=4$
When $b=6$,then $a=4,c=2$;When $b=6$,then $a=2,c=3$;When $b=6$,then $a=3,c=2$
When $b=6$,then $a=5,c=1$;When $b=6$,then $a=1,c=5$;When $b=6$,then $a=2,c=2$
When $b=6$,then $a=3,c=1$;When $b=6$,then $a=1,c=3$;When $b=6$,then $a=2,c=1$
When $b=6$,then $a=1,c=2$;When $b=6$,then $a=1,c=1$

I have counted 35 cases and my probability is $\frac{35}{216}$,but in the book $\frac{43}{216}$ is the answer.

Which 8 cases i have left,i checked but could not find.Please help me.

Solutions Collecting From Web of "Probability that the equation so formed will have real roots"

You seem to be missing the cases when $b= 5, 6$ and $a, c = 1$ and $c, a = 4, 6$.

It helps to have a partially-filled in multiplication table handy:

\begin{array}{c|cccccc}
& 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
1 & 1 & 2 & 3 & 4 & 5 & 6 \\
2 & 2 & 4 & 6 & 8 & & \\
3 & 3 & 6 & 9 & & & \\
4 & 4 & 8 & & & & \\
5 & 5 & & & & & \\
6 & 6 & & & & & \\
\end{array}

Then we can simply read off the pairs of numbers $(a,c)$ that satisfy
$ac \leq \frac14 b^2$ for each value of $b$:

\begin{align}
ac \leq \left\lfloor\tfrac14(2^2)\right\rfloor = 1:\qquad & (1,1) \\
ac \leq \left\lfloor\tfrac14(3^2)\right\rfloor = 2:\qquad & (1,1),(1,2),(2,1) \\
ac \leq \left\lfloor\tfrac14(4^2)\right\rfloor = 4:\qquad &
(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(3,1),(4,1) \\
ac \leq \left\lfloor\tfrac14(5^2)\right\rfloor = 6:\qquad &
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
(2,1),(2,2),(2,3), \\
& (3,1),(3,2), (4,1),(5,1),(6,1) \\
ac \leq \left\lfloor\tfrac14(6^2)\right\rfloor = 9:\qquad &
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
(2,1),(2,2),(2,3),(2,4) \\
& (3,1),(3,2),(3,3), (4,1),(4,2), (5,1),(6,1) \\
\end{align}

As you can see, there are $14$ pairs for $b=5$ and $17$ pairs for $b=6$,
but you have listed only $10$ rolls with $b=5$ and only $13$ with $b=6$.
That accounts for your eight missing rolls.
If you compare the lists you can discover which rolls you missed.

In the following $5$ cases equality holds:

$$\begin{matrix}
A B C\\
1 2 1\\
1 4 4\\
2 4 2\\
3 6 3\\
4 4 1\\
\end{matrix}$$

In $43$ cases the inequality holds. $43-5=38$, so you missed $3$ cases.

Ooops: I can see that Paul Sinclair has found the missing cases.