Question

12. Each coefficient in equation ax2 + bx + c = 0 is obtained by throwing a fair die. Find the
probability that the equation has equal roots.
statistics​

Answer 1

Given : Each coefficient in equation ax2 + bx + c = 0 is obtained by throwing a fair die

To Find : probability that the equation has equal roots.

Solution:

Equation ax² + bx + c = 0 has equal roots if

b²  - 4ac =  0  

b = 1 => 1  -  4ac  = 0   => 4ac = 1  not possible

b = 2  => 4 - 4ac = 0  => 4ac = 4 => ac  = 1

b = 2    ,   (a  , c) = (1 , 1)

b = 3 => 9 - 4ac  = 0 => 4ac = 9  => ac = 9/4  not possible

b = 4 => 16 - 4ac  = 0  => ac  = 4

b = 4     ,  (a , c)  = ( 1 , 4) , ( 2 , 2) , (  4, 1)

b = 5 => 25 - 4ac = 0  => 4ac = 25 => ac = 25/4  not possible

b = 6 => 36 - 4ac = 0  => ac  =  9

b = 6       (a , c)  = (  3 , 3)

Value of a , b & c can be 1 to 6

Total possible combination = 6 * 6 * 6  = 216

Combination when   Equation ax² + bx + c = 0 has equal roots  are 5

(a , b , c)  = (1  , 2 , 1) , (  1,  4 , 4) , ( 2 , 4 , 2 ) ( 4 , 4 , 1)  , ( 3 , 6 , 3)

Probability = 5/216

probability that the equation has equal roots.  = 5/216

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Answer 2

Answer = 5/216

Solution:-

Given - ax²+bx+c = 0 has equal roots i.e Real and equal

To Find - Probability That equation Has Equal roots

Solution/

b²-4ac = 0

If :-

I}b=1

(1)²-4ac —» 1-4ac—»4ac=1–»ac=1/4 (Impossible)

II}b=2

(2)²-4ac —» 4-4ac—»4ac=4—»ac=4/4=1(possible)

(If a and c are 1 and 1 respectively)

III}b=3

(3)²-4ac—»9-4ac—»4ac=9—»ac=9/4(Impossible)

Iv}b=4

(4)²-4ac—»16-4ac—»4ac=16—»ac=16/4=4(possible)

(If a and c are (1,4) (2,2) (4,1) respectively)

v}b=5

(5)²-4ac—»25-4ac—»4ac=25—»ac=25/4(Impossible)

vi}b=6

(6)²-4ac—»36-4ac—»4ac=36—»ac=36/4=9(Possible)

(If a and c are (3,3) respectively)

Maximum Values of A,B,C i.e Sample space (S) is 6*6*6 = 36*6= 216

Now,

(a,b,c) = {1,2,1 . 1,4,4 . 2,4,2 . 4,4,1 . 3,6,3}

n(a,b,c) = 5

Therefore, P(a,b,c) = n(a,b,c)/n(S)=

P(a,b,c) = 5/216 .