Aspire Faculty ID #3638 · Topic: NIMCET 2012 · Just now
NIMCET 2012

Coefficients a, b, c of $ax^2 + bx + c = 0$ are chosen by tossing 3 fair coins. Head means 1, Tail means 2. Find the probability that the roots are imaginary

Solution

Since each of $a,b,c$ can be $1$ or $2$: 
Total possible triples $(a,b,c)$: $2^3 = 8$. 
Roots are imaginary if: $b^2 - 4ac < 0$ 
So we check all $8$ cases. 

 List all possibilities 
 $(1,1,1)$: $1^2 - 4(1)(1) = 1 - 4 = -3 < 0$ ✔ 
 $(1,1,2)$: $1 - 8 = -7 < 0$ ✔ 
 $(1,2,1)$: $4 - 4 = 0$ ✘ (not imaginary) 
 $(1,2,2)$: $4 - 8 = -4 < 0$ ✔ 
 $(2,1,1)$: $1 - 8 = -7 < 0$ ✔ 
 $(2,1,2)$: $1 - 16 = -15 < 0$ ✔ 
 $(2,2,1)$: $4 - 8 = -4 < 0$ ✔ 
 $(2,2,2)$: $4 - 16 = -12 < 0$ ✔ 
 Count imaginary root cases 
Total imaginary cases = $7$ out of $8$. 
So probability: $\displaystyle P = \frac{7}{8}$ 
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