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Two fair dice are thrown. The numbers on them are taken as $\lambda$ and $\mu$, and a system of linear equations
$x+y+z=5$ ; $x+2 y+3 z=\mu$ ; $x+3 y+\lambda z=1$
is constructed. If $\mathrm{p}$ is the probability that the system has a unique solution and $\mathrm{q}$ is the probability that the system has no solution, then :
$\mathrm{p}=\frac{1}{6}$ and $\mathrm{q}=\frac{1}{36}$
$\mathrm{p}=\frac{5}{6}$ and $\mathrm{q}=\frac{5}{36}$
$\mathrm{p}=\frac{5}{6}$ and $\mathrm{q}=\frac{1}{36}$
$\mathrm{p}=\frac{1}{6}$ and $\mathrm{q}=\frac{5}{36}$
Solution
$\mathrm{D} \neq 0 \Rightarrow\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda\end{array}\right| \neq 0 \Rightarrow \lambda \neq 5$
For no solution $\mathrm{D}=0 \Rightarrow \lambda=5$
$\mathrm{D}_{1}=\left|\begin{array}{lll}1 & 1 & 5 \\ 1 & 2 & \mu \\ 1 & 3 & 1\end{array}\right| \neq 0 \Rightarrow \mu \neq 3$
$\mathrm{p}=\frac{5}{6}$
$\mathrm{q}=\frac{1}{6} \times \frac{5}{6}=\frac{5}{36}$
