Question

Two numbers $b$ and $c$ are chosen at random (with replacement) from the numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9. Find the probability that $x^2+bx+c+>0$ for all $x\in$ R

• A. $\frac{32}{81}$
• B. $\frac{59}{81}$
• C. $\frac{53}{81}$
• D. None of these
  

Answer 1

Answer: (A)

$\frac{32}{81}$

$x^2+bx+c>0$ for all $x\in R$
$\Rightarrow b^2-4c<0$
$b,c$ are choosen from $1$ to $9$ with replacement.
Total no of possibilities $=81$
Favourable outcomes are
$(1,9),(1,2).....(1,9)$ ------ 9 ways
$(2,2),(2,3)....(2,9)$ ------ 8 ways
$(3,3),(3,4)....(3,9)$ ------ 7 ways
$(4,5),(4,6).....(4,9)$ ----- 5 ways
$(5,7),(5,8),(5,9)$ ----- 3 ways
favourable outcomes = $32$
$\therefore$ Probability $=\frac{32}{81}$