Probability Ques 10
Consider the system of equations
$ a x+b y=0, c x+d y=0 $
$ \text { where } \quad a, b, c, d \in\{0,1\} \text {. } $
Statement I The probability that the system of equations has a unique solution, is $3 / 8$.
Statement II The probability that the system of equations has a solution, is 1 .
$(2008,3 M)$
Show Answer
Answer:
Correct Answer: 10.(b)
Solution:
Formula:
- The number of all possible determinants of the form
$ \left|\begin{array}{ll} a & b \\ c & d \end{array}\right|=2^{4}=16 $
Out of which only 10 determinants given by
$ \begin{aligned} & \left|\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right|,\left|\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right|,\left|\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right|,\left|\begin{array}{ll} 0 & 0 \\ 1 & 1 \end{array}\right|,\left|\begin{array}{ll} 0 & 1 \\ 0 & 1 \end{array}\right|,\left|\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}\right|,\left|\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right|, \\ & \left|\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right|,\left|\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right| \end{aligned} $
Vanish and remaining six determinants have non-zero values. Hence, the required probability $=\frac{6}{16}=\frac{3}{8}$
Statement I is true.
Statement II is also true as the homogeneous equations have always a solution and Statement II is not the correct explanation of Statement I.
BETA
