If the numbers appeared on the two throws of a fair six faced die are α and β, then the probability that x2+α x+β0, for all x ∈ R, is :

Question

If the numbers appeared on the two throws of a fair six faced die are α and β, then the probability that x2+α x+β>0, for all x ∈ R, is : (A) (17/36) (B) (4/9) (C) (1/2) (D) (19/36)

Tardigrade Admin · Accepted Answer

x 2+α x +β>0, ∀ x ∈ R D =α2-4 β<0 α2<4 β Total cases = 6 × 6=36 Fav. cases =β=1, α=1 β=2, α=1,2 β=3, α=1,2,3 β=4, α=1,2,3 β=5, α=1,2,3,4 β=6, α=1,2,3,4 Total favourable cases =17 P ( x )=(17/36)

Q. If the numbers appeared on the two throws of a fair six faced die are $α$ and $β$ , then the probability that $x^{2} + αx + β > 0$ , for all $x \in R$ , is :

$\frac{17}{36}$

$\frac{4}{9}$

$\frac{1}{2}$

$\frac{19}{36}$

Q. If the numbers appeared on the two throws of a fair six faced die are α and β, then the probability that x2+αx+β>0, for all x∈R, is :

3617​

94​

21​

3619​

x2+αx+β>0,∀x∈R D=α2−4β<0 α2<4β Total cases=6×6=36 Fav. cases =β=1,α=1 β=2,α=1,2 β=3,α=1,2,3 β=4,α=1,2,3 β=5,α=1,2,3,4 β=6,α=1,2,3,4 Total favourable cases =17 P(x)=3617​

Q. If the numbers appeared on the two throws of a fair six faced die are $α$ and $β$ , then the probability that $x^{2} + αx + β > 0$ , for all $x \in R$ , is :

$\frac{17}{36}$

$\frac{4}{9}$

$\frac{1}{2}$

$\frac{19}{36}$