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← IIT-JEE 2008 · Paper 1
MathematicsAssertion and reason+3 / -1Question 14 of 69

IIT-JEE 2008 · Paper 1, Mathematics question 14

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\dfrac38.

and

STATEMENT-2: The probability that the system of equations has a solution is 1.

Choose one option
Answer and worked solution

Official answer: B

Statement 1. Of the 16 choices of (a,b,c,d)∈{0,1}4(a,b,c,d)\in\{0,1\}^4, a unique solution needs ad−bc≠0ad-bc\ne0:

  • ad=1ad=1, bc=0bc=0: a=d=1a=d=1 with (b,c)≠(1,1)(b,c)\ne(1,1): 3 cases;
  • ad=0ad=0, bc=1bc=1: b=c=1b=c=1 with (a,d)≠(1,1)(a,d)\ne(1,1): 3 cases.

Probability 616=38\frac6{16}=\frac38. True.

Statement 2. A homogeneous system always has the solution x=y=0x=y=0. True.

Explanation? Statement 2 says nothing about uniqueness, so it does not explain Statement 1.

(B)