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Probability of Unique Solution in Binary Linear System

Consider the system of equations
ax+by=0ax + by = 0, cx+dy=0cx + dy = 0, where a,b,c,d∈{0,1}a, b, c, d \in \{0, 1\} .
STATEMENT-1 : The probability that the system of equations has a unique solution is 38\frac{3}{8} .
and
STATEMENT-2 : The probability that the system of equations has a solution is 11 .

Options

A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

Correct
C

STATEMENT-1 is True, STATEMENT-2 is False

D

STATEMENT-1 is False, STATEMENT-2 is True

Step-by-Step Solution

To determine the correctness of the given statements, we analyze the homogeneous system of linear equations:

{ax+by=0cx+dy=0\begin{cases} ax + by = 0 \\ cx + dy = 0 \end{cases}

where each coefficient a,b,c,d∈{0,1}a, b, c, d \in \{0, 1\}.

1. Analysis of STATEMENT-2:

The given system is a homogeneous system of linear equations.
For any values of a,b,c,d∈{0,1}a, b, c, d \in \{0, 1\}, the trivial solution (x,y)=(0,0)(x, y) = (0, 0) always satisfies both equations:

a(0)+b(0)=0andc(0)+d(0)=0a(0) + b(0) = 0 \quad \text{and} \quad c(0) + d(0) = 0

Since a solution always exists for all possible choices of (a,b,c,d)(a, b, c, d), the probability that the system of equations has a solution is:

P(system has a solution)=1P(\text{system has a solution}) = 1

Therefore, STATEMENT-2 is True.


2. Analysis of STATEMENT-1:

The total number of possible ordered quadruplets (a,b,c,d)(a, b, c, d) is:

2×2×2×2=24=162 \times 2 \times 2 \times 2 = 2^4 = 16

A homogeneous system of two linear equations in two variables has a unique solution (the trivial solution) if and only if the determinant of its coefficient matrix is non-zero:

Δ=∣abcd∣=ad−bc≠0\Delta = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \neq 0

Since a,b,c,d∈{0,1}a, b, c, d \in \{0, 1\}, the products adad and bcbc can each only take the values 00 or 11. Thus, Δ≠0\Delta \neq 0 can occur in two disjoint cases:

  1. Case 1: ad=1ad = 1 and bc=0bc = 0

    • ad=1  ⟹  a=1ad = 1 \implies a = 1 and d=1d = 1 (1 choice).
    • bc=0  ⟹  (b,c)∈{(0,0),(0,1),(1,0)}bc = 0 \implies (b, c) \in \{(0, 0), (0, 1), (1, 0)\} (3 choices).
    • Number of outcomes =1×3=3= 1 \times 3 = 3.
  2. Case 2: ad=0ad = 0 and bc=1bc = 1

    • bc=1  ⟹  b=1bc = 1 \implies b = 1 and c=1c = 1 (1 choice).
    • ad=0  ⟹  (a,d)∈{(0,0),(0,1),(1,0)}ad = 0 \implies (a, d) \in \{(0, 0), (0, 1), (1, 0)\} (3 choices).
    • Number of outcomes =3×1=3= 3 \times 1 = 3.

Thus, the total number of favorable outcomes for a unique solution is:

n(E)=3+3=6n(E) = 3 + 3 = 6

The probability that the system has a unique solution is:

P(unique solution)=n(E)n(S)=616=38P(\text{unique solution}) = \frac{n(E)}{n(S)} = \frac{6}{16} = \frac{3}{8}

Therefore, STATEMENT-1 is True.


3. Relationship between the Statements:

STATEMENT-2 is a property of any homogeneous linear system (the existence of the trivial solution), but it does not account for the condition Δ≠0\Delta \neq 0 which determines whether the system has a unique solution or infinitely many solutions. Hence, STATEMENT-2 is not a correct explanation for STATEMENT-1.

Therefore, the correct choice is B.