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(B) For the quadratic equation $x^2+ax+b=0$ to have real roots, the discriminant $D$ must be greater than or equal to $0$. $D = a^2 - 4b \geq 0 \impli

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$\frac{5}{6}$

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(B) For the quadratic equation $x^2+ax+b=0$ to have real roots, the discriminant $D$ must be greater than or equal to $0$. $D = a^2 - 4b \geq 0 \implies a^2 \geq 4b$. Given $a \in \{3, 4, 5\}$ and $b \in \{1, 2, 3, 4\}$, the total number of possible pairs $(a, b)$ is $3 	imes 4 = 12$. We check the condition $a^2 \geq 4b$ for each $a$: $1$. If $a=3$, $a^2=9$. $9 \geq 4b \implies b \leq 2.25$. Possible $b$ values are $\{1, 2\}$ ($2$ pairs). $2$. If $a=4$, $a^2=16$. $16 \geq 4b \implies b \leq 4$. Possible $b$ values are $\{1, 2, 3, 4\}$ ($4$ pairs). $3$. If $a=5$, $a^2=25$. $25 \geq 4b \implies b \leq 6.25$. Possible $b$ values are $\{1, 2, 3, 4\}$ ($4$ pairs). Total favorable outcomes = $2 + 4 + 4 = 10$. Probability = $\frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{10}{12} = \frac{5}{6}$.

If the coefficients $a$ and $b$ of a quadratic expression $x^2+ax+b$ are chosen from the sets $A=\{3,4,5\}$ and $B=\{1,2,3,4\}$ respectively, then the probability that the equation $x^2+ax+b=0$ has real roots is

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