Each coefficient in the equation $a x^{2} + b x + c = 0$ is determined by throwing an ordinary die. The probability that the equation will have real roots is $\frac{7 k + 1}{216}$ . Find the value of $k$ .

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Solution

Let A denote the event that ${a x}^{2} + b x + c = 0$ has non-real complex roots. Then $¯ ¯¯ ¯ A$ is the event that $a x^{2} + b x + c = 0$ has real roots. We have,
$P (¯ ¯¯ ¯ A) = 1 - P (A)$
${a x}^{2} + b x + c = 0$ has real roots if and only if $b^{2} - 4 a c \geq 0$ or $b^{2} \geq 4 a c$ . Total number of cases $= 6 \times 6 \times 6 = 216.$
We now proceed to find favorable cases. Since maximum value of b is 6, $b^{2} \leq 36.$ Therefore, $4 a c \leq 36$ or $a c \leq 9$
$a c$ $4 a c$ $(a, c)$
$b$ no. of cases
$1$ $4$ (1,1) 2,3,4,5,6, 5
$2$ $8$ (1,2),(2,1) 3,4,5,6, 8
$3$ $12$ (1,3),(3,1) 4,5,6 6
$4$ $16$ (1,4),(2,2),(4,1) 4,5,6 9
$5$ $20$ (5,1),(5,1) 5,6 4
$6$ $24$ (1,6),(2,3),(3,2),(6,1) 5,6 8
$7$ $28$ - - 0
$8$ $32$ (4,2),(2,4) 6 2
$9$ $36$ (3,3) 6 1
Thus, total number of Cases
$5 + 8 + 6 + 9 + 4 + 8 + 2 + 1 = 43$
Thus, $P (¯ ¯¯ ¯ A) = \frac{43}{216}$
hence $k = 6$

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$a c$ $4 a c$	$(a, c)$		$b$	no. of cases
$1$	$4$	(1,1)	2,3,4,5,6,	5
$2$	$8$	(1,2),(2,1)	3,4,5,6,	8
$3$	$12$	(1,3),(3,1)	4,5,6	6
$4$	$16$	(1,4),(2,2),(4,1)	4,5,6	9
$5$	$20$	(5,1),(5,1)	5,6	4
$6$	$24$	(1,6),(2,3),(3,2),(6,1)	5,6	8
$7$	$28$	-	-	0
$8$	$32$	(4,2),(2,4)	6	2
$9$	$36$	(3,3)	6	1

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