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Grade: 11
        
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 
3 years ago

Answers : (3)

deepak
84 Points
							
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
 
3 years ago
deepak
84 Points
							
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
3 years ago
deepak
84 Points
							
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 instead of
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
3 years ago
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