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        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

deepak
84 Points
							if x2+bx+c>0then it implies that discriminant should be less than zeroi.e. b2-4c>0hence b>2$\sqrt{c}$ if c=1 then b can be 3,4,5,6,7,8,9  …..7if c=2 then b can be 3,4,5,6,7,8,9  …....7if c=3 then b can be 4,5,6,7,8,9 ........6if c=4 then b can be 5,6,7,8,9 …........5if c=5 then b can be 5,6,7,8,9 …........5if c=6 then b can be 5,6,7,8,9 ….…...5if c=7 then b can be 6,7,8,9 ….….4if c=8 then b can be 6,7,8,9..........4if 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>0then it implies that discriminant should be less than zero  i.e. b2-4c’hence b$\sqrt{c}$ if c=1 then b can be 1if c=2 then b can be 1,2if c=3 then b can be 1,2,3if c=4 then b can be 1,2,3if c=5 then b can be 1,2,3,4if c=6 then b can be 1,2,3,4if c=7 then b can be 1,2,3,4,5if c=8 then b can be 1,2,3,4,5if 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>0then 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 1if c=2 then b can be 1,2if c=3 then b can be 1,2,3if c=4 then b can be 1,2,3if c=5 then b can be 1,2,3,4if c=6 then b can be 1,2,3,4if c=7 then b can be 1,2,3,4,5if c=8 then b can be 1,2,3,4,5if 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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