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(C) The set $S$ consists of equations $x^2+bx+c=0$ where $b, c \in \{1, 2, 3, 4, 5, 6\}$.The total number of such equations is $n(S) = 6 	imes 6 = 36

Vedclass · Accepted Answer

$\frac{19}{36}$

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(C) The set $S$ consists of equations $x^2+bx+c=0$ where $b, c \in \{1, 2, 3, 4, 5, 6\}$.The total number of such equations is $n(S) = 6 	imes 6 = 36$.$A$ quadratic equation $x^2+bx+c=0$ has real roots if the discriminant $D = b^2-4ac \geq 0$.Since $a=1$,the condition becomes $b^2 \geq 4c$.We test values of $b$ from $1$ to $6$:If $b=1$,$1 \geq 4c$ (No values for $c$)If $b=2$,$4 \geq 4c \implies c \leq 1$ ($c=1$,$1$ case)If $b=3$,$9 \geq 4c \implies c \leq 2.25$ ($c=1, 2$,$2$ cases)If $b=4$,$16 \geq 4c \implies c \leq 4$ ($c=1, 2, 3, 4$,$4$ cases)If $b=5$,$25 \geq 4c \implies c \leq 6.25$ ($c=1, 2, 3, 4, 5, 6$,$6$ cases)If $b=6$,$36 \geq 4c \implies c \leq 9$ ($c=1, 2, 3, 4, 5, 6$,$6$ cases)Total favorable outcomes $n(E) = 0 + 1 + 2 + 4 + 6 + 6 = 19$.The probability $P(E) = \frac{n(E)}{n(S)} = \frac{19}{36}$.

Let $S$ be the set of all quadratic equations of the form $x^2+bx+c=0$,where $b, c \in \{1, 2, 3, 4, 5, 6\}$. If an equation is selected at random from $S$,then the probability that the equation has real roots is

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