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. \(p = \frac{1}{6}\) and \(q = \frac{1}{36}\)
• B. \(p = \frac{5}{6}\) and \(q = \frac{1}{36}\)
• C. \(p = \frac{1}{6}\) and \(q = \frac{5}{36}\)
• D. \(p = \frac{5}{6}\) and \(q = \frac{5}{36}\)

Hint:

For problems involving systems of equations and probability, first establish the algebraic conditions on the parameters using determinants. Then, translate these conditions into counting the number of favorable outcomes based on the constraints (like dice rolls).

Answer 1

The Correct Option is D

Step 1: Understanding the Question 
We are given a system of linear equations with parameters \(\lambda\) and \(\mu\) determined by rolling two dice. We need to find the probabilities of the system having a unique solution (\(p\)) and no solution (\(q\)). 

Step 2: Key Formula or Approach 
The nature of the solution of a system of linear equations \(AX=B\) is determined by the determinant of the coefficient matrix, \(D = \det(A)\), and the determinants \(D_x, D_y, D_z\).

Unique Solution: \(D \neq 0\).
No Solution: \(D = 0\) and at least one of \(D_x, D_y, D_z\) is non-zero.
Infinite Solutions: \(D = D_x = D_y = D_z = 0\).

Step 3: Detailed Explanation 
Calculate the determinant D: 
 

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{vmatrix} = 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2) = 2\lambda - 9 - \lambda + 3 + 1 = \lambda - 5 \]

Calculate Probability p (Unique Solution): 
For a unique solution, \(D \neq 0\), which means \(\lambda - 5 \neq 0\), so \(\lambda \neq 5\).
Since \(\lambda\) is the outcome of a fair die, \(\lambda \in \{1, 2, 3, 4, 5, 6\}\).
The favorable outcomes for \(\lambda\) are \{1, 2, 3, 4, 6\}. There are 5 favorable outcomes.
The value of \(\mu\) can be anything. Total outcomes for rolling two dice = \(6 \times 6 = 36\). Favorable outcomes = \(5 \times 6 = 30\).
 

\[ p = P(\lambda \neq 5) = \frac{30}{36} = \frac{5}{6} \]

Calculate Probability q (No Solution): 
For no solution, we need \(D = 0\), which means \(\lambda = 5\).
Additionally, at least one of \(D_x, D_y, D_z\) must be non-zero. Let's calculate \(D_z\).
 

\[ D_z = \begin{vmatrix} 1 & 1 & 5 \\ 1 & 2 & \mu \\ 1 & 3 & 1 \end{vmatrix} = 1(2 - 3\mu) - 1(1 - \mu) + 5(3 - 2) = 2 - 3\mu - 1 + \mu + 5 = 6 - 2\mu \]

For no solution, we need \(D=0\) and \(D_z \neq 0\). \(\lambda = 5\) and \(6 - 2\mu \neq 0 \implies 2\mu \neq 6 \implies \mu \neq 3\).
So, the condition for no solution is \(\lambda = 5\) and \(\mu \neq 3\).
\(\mu\) can be 1, 2, 4, 5, 6. There are 5 favorable values for \(\mu\).
The favorable pairs \((\lambda, \mu)\) are (5,1), (5,2), (5,4), (5,5), (5,6). There are 5 such pairs.
Total possible pairs = 36.
 

\[ q = \frac{5}{36} \]

Step 4: Final Answer 
The probabilities are \(p = \frac{5}{6}\) and \(q = \frac{5}{36}\).