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(B) The given equation is $ax^{2} + bx + 1 = 0$.For the equation to have real roots, the discriminant $D$ must be greater than or equal to $0$.$D = b^

Vedclass · Accepted Answer

$1/4$

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(B) The given equation is $ax^{2} + bx + 1 = 0$.For the equation to have real roots, the discriminant $D$ must be greater than or equal to $0$.$D = b^{2} - 4ac \geq 0$.Since $c = 1$, we have $b^{2} - 4a \geq 0$.The possible values for $(a, b)$ are $(1, 1), (1, 2), (2, 1), (2, 2)$, each with a probability of $1/4$.Checking the condition $b^{2} - 4a \geq 0$ for each pair:$1$. For $(1, 1): 1^{2} - 4(1) = -3 $2$. For $(1, 2): 2^{2} - 4(1) = 0 \geq 0$ (Real).$3$. For $(2, 1): 1^{2} - 4(2) = -7 $4$. For $(2, 2): 2^{2} - 4(2) = -4 Only the pair $(1, 2)$ satisfies the condition.Thus, the required probability is $1/4$.

Each of $a$ and $b$ can take values $1$ or $2$ with equal probability. The probability that the equation $ax^{2} + bx + 1 = 0$ has real roots is equal to

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