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:

We have two independent fair dice. Each die can show any integer from $$1$$ to $$6$$ with equal probability $$\displaystyle \frac16$$. The first die gives the value $$\lambda$$ and the second die gives the value $$\mu$$. Using these values the following system of linear equations is formed:

$$$ \begin{aligned} x + y + z &= 5,\\[4pt] x + 2y + 3z &= \mu,\\[4pt] x + 3y + \lambda z &= 1. \end{aligned} $$$

For a system of three linear equations in the three unknowns $$x,\,y,\,z$$, the basic fact is:

• If the determinant of the coefficient matrix is non-zero, the system has a unique solution.
• If the determinant is zero, the system may be either consistent (infinitely many solutions) or inconsistent (no solution). Consistency has to be checked separately through the augmented matrix.

First we write the coefficient matrix $$A$$ and find its determinant. The matrix is

$$$ A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 2 & 3\\ 1 & 3 & \lambda \end{bmatrix}, $$$

and its determinant is

$$$ \begin{aligned} \det A &=\;1 \Bigl(2\lambda-3\!\cdot\!3\Bigr) \;-\;1 \Bigl(1\!\cdot\!\lambda-3\!\cdot\!1\Bigr) \;+\;1 \Bigl(1\!\cdot\!3-2\!\cdot\!1\Bigr)\\[6pt] &=\;1(2\lambda-9)\;-\;(\lambda-3)\;+\;1\\[6pt] &=\;2\lambda-9-\lambda+3+1\\[6pt] &=\;\lambda-5. \end{aligned} $$$

So $$\det A=\lambda-5$$.

Now, $$\det A\neq0 \iff \lambda\neq5$$. Because $$\lambda$$ comes from a fair die, five of the six possible values (namely $$1,2,3,4,6$$) satisfy $$\lambda\neq5$$. Therefore

$$$ p = P(\text{unique solution}) = P(\lambda\neq5) = \frac{5}{6}. $$$

Next we look at the case $$\lambda=5$$, where the determinant is zero. We must decide when the system is inconsistent.

Setting $$\lambda=5$$, the equations become

$$$ \begin{aligned} x + y + z &= 5, \quad -(1)\\[4pt] x + 2y + 3z &= \mu, \quad -(2)\\[4pt] x + 3y + 5z &= 1. \quad -(3) \end{aligned} $$$

We eliminate $$x$$ by subtracting equation (1) from equations (2) and (3):

$$$ \begin{aligned} \text{(2)}-\text{(1)}&:;;0x + y + 2z = \mu - 5,\\[4pt] \text{(3)}-\text{(1)}&:;;0x + 2y + 4z = 1 - 5 = -4. \end{aligned} $$$

The left-hand sides satisfy $$[0,2,4]=2[0,1,2]$$, so the two new equations are multiples of each other. For the system to be consistent, the right-hand sides must have the same ratio; that is, we need

$$$ -4 = 2(\mu - 5)\;\Longrightarrow\;-4 = 2\mu - 10\;\Longrightarrow\;2\mu = 6\;\Longrightarrow\;\mu = 3. $$$

Thus:

• If $$\lambda=5$$ and $$\mu=3$$, the system is consistent (in fact it has infinitely many solutions).
• If $$\lambda=5$$ and $$\mu\neq3$$, the system is inconsistent and hence has no solution.

The probability of inconsistency is therefore

$$$ q = P(\lambda=5\ \text{and}\ \mu\neq3) = P(\lambda=5)\cdot P(\mu\neq3) = \frac16 \times \frac56 = \frac{5}{36}. $$$

We have found $$p=\dfrac56$$ and $$q=\dfrac{5}{36}$$, which corresponds to option D.

Hence, the correct answer is Option D.