two numbers b and c are chosen at random with replacement from the numbers 1,2,3,4,5,6,7,8,9 .find the probability that x2+bx+c>0 for all x belongs to R

deepak
84 Points
5 years ago
if x2+bx+c>0
then it implies that discriminant should be less than zero
i.e. b2-4c>0
hence b>2$\sqrt{c}$

if c=1 then b can be 3,4,5,6,7,8,9  …..7
if c=2 then b can be 3,4,5,6,7,8,9  …....7
if c=3 then b can be 4,5,6,7,8,9 ........6
if c=4 then b can be 5,6,7,8,9 …........5
if c=5 then b can be 5,6,7,8,9 …........5
if c=6 then b can be 5,6,7,8,9 ….…...5
if c=7 then b can be 6,7,8,9 ….….4
if c=8 then b can be 6,7,8,9..........4
if c=9 then b can be 7,8,9 ….3

therefore total no of favourable pairs are 46

then the probability of the given even happening is 46/81 = 0.567

hope it is useful

deepak
84 Points
5 years ago
sorry that is the incorrect solution this is the correct one
if x2+bx+c>0
then it implies that discriminant should be less than zero
i.e. b2-4c’
hence b$\sqrt{c}$

if c=1 then b can be 1
if c=2 then b can be 1,2
if c=3 then b can be 1,2,3
if c=4 then b can be 1,2,3
if c=5 then b can be 1,2,3,4
if c=6 then b can be 1,2,3,4
if c=7 then b can be 1,2,3,4,5
if c=8 then b can be 1,2,3,4,5
if c=9 then b can be 1,2,3,4,5

therefore total no of favourable pairs are 32

then the probability of the given even happening is 32/81 = 0.395

hope it was useful
deepak
84 Points
5 years ago
sorry that is the incorrect solution this is the correct one
if x2+bx+c>0
then it implies that discriminant should be less than zero
hence b$\sqrt{c}$

if c=1 then b can be 1
if c=2 then b can be 1,2
if c=3 then b can be 1,2,3
if c=4 then b can be 1,2,3
if c=5 then b can be 1,2,3,4
if c=6 then b can be 1,2,3,4
if c=7 then b can be 1,2,3,4,5
if c=8 then b can be 1,2,3,4,5
if c=9 then b can be 1,2,3,4,5

therefore total no of favourable pairs are 32

then the probability of the given even happening is 32/81 = 0.395

hope it was useful