Probability that the equation so formed will have real roots

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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.