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If the coefficients $a$ and $b$ of $a$ quadratic expression $x^2+a x+b$ are choosen from the sets $A=\{3,4,5\}$ and $\mathrm{B}=\{1,2,3,4\}$ respectively, then the probability that the equation $x^2+a x+b=0$ has real roots is
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The correct answer is:
$\frac{5}{6}$
$x^2+a x+b=0$
$\begin{aligned} & \mathrm{D} \geq 0 \\ & a^2-4 b \geq 0 \\ & a=3, \quad b=1,2 \\ & a=4, \quad b=1,2,3,4 \\ & a=5, \quad b=1,2,3,4 \\ & \text { Possible pairs }=10 \\ & \text { Total pairs }=12 \\ & \text { Probability }=\frac{10}{12}=\frac{5}{6}\end{aligned}$
$\begin{aligned} & \mathrm{D} \geq 0 \\ & a^2-4 b \geq 0 \\ & a=3, \quad b=1,2 \\ & a=4, \quad b=1,2,3,4 \\ & a=5, \quad b=1,2,3,4 \\ & \text { Possible pairs }=10 \\ & \text { Total pairs }=12 \\ & \text { Probability }=\frac{10}{12}=\frac{5}{6}\end{aligned}$
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