Two fair dice are thrown.

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 :

Accepted Answer

$$D \ne 0 \Rightarrow \left| {\begin{pmatrix} 1 & 1 & 1  \
   1 & 2 & 3  \
   1 & 3 & \lambda   \

\end{pmatrix} } \right| \ne 0 \Rightarrow \lambda  \ne 5$$

For no solution D = 0 $$\Rightarrow$$ $$\lambda$$ = 5

$${D_1} = \left| {\begin{pmatrix} 1 & 1 & 5  \
   1 & 2 & \mu   \
   1 & 3 & 1  \

\end{pmatrix} } \right| \ne 0 \Rightarrow \mu  \ne 3$$

$$p = {5 \over 6}$$

$$q = {1 \over 6} \times {5 \over 6} = {5 \over {36}}$$

Option (b).

Solutions · 1