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 Consider the system of equations ax + by = 0, cx + dy = 0, where a, b, c, d ∈ {0, 1}
Statement-1 : The probability that the system of equations has a unique solution is 3/8.
Statement-2 : The probability that the system of equations has a solution is 1.
  • 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.
  • c)
    Statement-1 is true, Statement-2 is false.
  • d)
    Statement-1 is false, Statement-2 is true.
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Consider the system of equations ax + by = 0, cx + dy = 0, where a, b,...
Favourable number fo caes is 6 (either ad = 1, bc =0 or ad = 0, bc =1). The probability that system of equation has unique solution is 6/16 = 3/8. Since x = 0 satisfies both the equations, the system has at least one solution.
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Most Upvoted Answer
Consider the system of equations ax + by = 0, cx + dy = 0, where a, b,...
Given:
System of equations: ax + by = 0, cx + dy = 0
a, b, c, d ∈ {0, 1}

To find:
The probability that the system of equations has a unique solution.

Solution:
Let's analyze each statement separately.

Statement-1: The probability that the system of equations has a unique solution is 3/8.

To find the probability of a unique solution, we need to analyze the different cases that can occur.

Case 1: a = 0, b = 0, c = 0, d = 0
In this case, both equations become 0x + 0y = 0, which is always true. There are infinitely many solutions in this case.

Case 2: a = 1, b = 1, c = 1, d = 1
In this case, both equations become 1x + 1y = 0 and 1x + 1y = 0, which are equivalent equations. There are infinitely many solutions in this case.

Case 3: a = 0, b = 0, c = 1, d = 1
In this case, the first equation becomes 0x + 0y = 0, which is always true. The second equation becomes 1x + 1y = 0, which has a unique solution (x = 0, y = 0).

Case 4: a = 1, b = 0, c = 0, d = 1
In this case, the first equation becomes 1x + 0y = 0, which has a unique solution (x = 0, y = 0). The second equation becomes 0x + 1y = 0, which also has a unique solution (x = 0, y = 0).

Case 5: All other possible combinations of a, b, c, d
In these cases, either one of the equations becomes 0x + 0y = 0 or both equations become equivalent equations (1x + 1y = 0). In both cases, there are infinitely many solutions.

Out of the 16 possible combinations of a, b, c, d, only 2 cases have a unique solution (Case 3 and Case 4). Therefore, the probability of a unique solution is 2/16 = 1/8.

Hence, Statement-1 is false.

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

The system of equations will always have a solution because even in the worst-case scenario (Case 1 where a = 0, b = 0, c = 0, d = 0), the equations are always satisfied. Therefore, Statement-2 is true.

Conclusion:
Statement-1 is false, and Statement-2 is true. However, Statement-2 is not a correct explanation for Statement-1.

Therefore, the correct answer is option 'B'.
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Consider the system of equations ax + by = 0, cx + dy = 0, where a, b, c, d ∈ {0, 1}Statement-1 : The probability that the system of equations has a unique solution is 3/8.Statement-2 : The probability that the system of equations has a solution is 1.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.c)Statement-1 is true, Statement-2 is false.d)Statement-1 is false, Statement-2 is true.Correct answer is option 'B'. Can you explain this answer?
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