TS EAMCET · Maths · Probability
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
- A \(\frac{1}{6}\)
- B \(\frac{5}{6}\)
- C \(\frac{3}{4}\)
- D \(\frac{7}{12}\)
Answer & Solution
Correct Answer
(B) \(\frac{5}{6}\)
Step-by-step Solution
Detailed explanation
Given equation, \(x^2+a x+b\) has real roots if \(a^2 \geq 4 b\) favourable outcomes is \(\begin{aligned} & (3,1)(3,2)(4,1)(4,2)(4,3),(4,4) \\ & (5,1)(5,2)(5,3)(5,4) \\ & \therefore \text { Required probability }=\frac{10}{3 \times 4}=\frac{10}{12}=\frac{5}{6} \end{aligned}\)
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