Question

If the numbers appeared on the two throws of a fair six faced die are α and β, then the probability that x 2 + α x + β> 0, for all _x _∈ R , is :

• A. $\frac{17}{36}$
• B. $\frac{4}{9}$
• C. $\frac{1}{2}$
• D. $\frac{19}{36}$

Answer 1

Answer: (A)

$\frac{17}{36}$

Answer 2

For x 2 + α x + β> 0 ∀ x ∈ R to hold, we should have α2 – 4β<0
If α = 1, β can be 1, 2, 3, 4, 5, 6 i.e. , 6 choices
If α = 2, β can be 2, 3, 4, 5, 6 i.e. , 5 choices
If α = 3, β can be 3, 4, 5, 6 i.e. , 4 choices
If α = 4, β can be 5 or 6 i.e. , 2 choices
If α = 6, No possible value for β i.e. , 0 choices
Hence total favourable outcomes
= 6 + 5 + 4 + 2 + 0 + 0
= 17
Total possible choices for α and β = 6 × 6 = 36
Required probability = $\frac{17}{36}$