Question

Two fair dice are thrown. The numbers on them are taken as $\lambda$ and $\mu ,$ and a system of linear equations
$x+y+z=5$
$x+2y+3z=\mu$
$x+3y+\lambda z=1$
is constructed. If $p$ is the probability that the system has a unique solution and $q$ is the probability that the system has no solution, then

• A. $p=\frac{1}{6}\text{and}q=\frac{1}{36}$
• B. $p=\frac{5}{6}\text{and}q=\frac{5}{36}$
• C. $p=\frac{1}{6}\text{and}q=\frac{5}{36}$
• D. $p=\frac{5}{6}\text{and}q=\frac{1}{36}$

Answer 1

The correct option is B $p=\frac{5}{6}\text{and}q=\frac{5}{36}$

$\left|\begin{array}{ccc}1& 3& \lambda \\ 1& 1& 1\\ 1& 2& 3\end{array}\right|=0\Rightarrow 1-3\left(2\right)+\lambda \left(1\right)=0\Rightarrow \lambda =5$

For $\lambda \ne 5$ there will be unique solution
$p=1-\frac{1}{6}=\frac{5}{6}$

For $\lambda =5$ and $\mu =3$ there will be infinitely many solutions and for $\lambda =5$ and $\mu \ne 3$ there will be no soluiton.

$q=\frac{1}{6}\cdot (1-\frac{1}{6})=\frac{5}{36}$