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If the coefficients $a$ and $b$ of a quadratic expression $x^2+a x+b$ are chosen from the sets $A=\{3,4,5\}$ and $B=\{1,2,3,4\}$ respectively, then the probability that the equation $x^2+a x+b=0$ has real roots is
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The correct answer is:
$\frac{5}{6}$
Given equation,
$x^2+a x+b$ has real roots if $a^2 \geq 4 b$ favourable outcomes is
$\begin{aligned}
& (3,1)(3,2)(4,1)(4,2)(4,3),(4,4) \\
& (5,1)(5,2)(5,3)(5,4) \\
& \therefore \text { Required probability }=\frac{10}{3 \times 4}=\frac{10}{12}=\frac{5}{6}
\end{aligned}$
$x^2+a x+b$ has real roots if $a^2 \geq 4 b$ favourable outcomes is
$\begin{aligned}
& (3,1)(3,2)(4,1)(4,2)(4,3),(4,4) \\
& (5,1)(5,2)(5,3)(5,4) \\
& \therefore \text { Required probability }=\frac{10}{3 \times 4}=\frac{10}{12}=\frac{5}{6}
\end{aligned}$
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