Two fair dice are thrown. The numbers on them are taken as λ and µ, and a system of linear equations
$x + y + z = 5, x + 2 y + 3 z = μ and x + 3 y + λz = 1$
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

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$p = \frac{1}{6} and q = \frac{1}{36}$

$p = \frac{5}{6} and q = \frac{5}{36}$

$p = \frac{5}{6} and q = \frac{1}{36}$

$p = \frac{1}{6} and q = \frac{5}{36}$

answer is B.

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Detailed Solution

$\begin{array}{l} x + y + z = 5 \\ x + 2 y + 3 z = μ \\ x + 3 y + λz = 1 \\ Δ = |\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & λ \end{array}| = (2 λ - 9) + (3 - λ) + (3 - 2) = λ - 5 \\ For unique solution Δ \neq 0 \\ \Rightarrow λ \neq 5 \\ And Δ_{1} or Δ_{2} or Δ_{3} \neq 0 \\ Δ_{3} = |\begin{array}{lll} 1 & 1 & 5 \\ 1 & 2 & μ \\ 1 & 3 & 1 \end{array}| = (2 - 3 μ) + (μ - 1) + 5 = 6 - 2 μ \\ If Δ_{3} \neq 3 and Δ = 0, then no solution \\ μ \neq 3 and λ = 5 \\ p = Probability of unique solution = \frac{5}{6} \\ q = Probability of no solution = \frac{5}{6} \cdot \frac{1}{6} = \frac{5}{36} \end{array}$