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 :
Solution
$$D \ne 0 \Rightarrow \left| {\matrix{
1 & 1 & 1 \cr
1 & 2 & 3 \cr
1 & 3 & \lambda \cr
} } \right| \ne 0 \Rightarrow \lambda \ne 5$$<br><br>For no solution D = 0 $\Rightarrow$ $\lambda$ = 5<br><br>$${D_1} = \left| {\matrix{
1 & 1 & 5 \cr
1 & 2 & \mu \cr
1 & 3 & 1 \cr
} } \right| \ne 0 \Rightarrow \mu \ne 3$$<br><br>$p = {5 \over 6}$<br><br>$q = {1 \over 6} \times {5 \over 6} = {5 \over {36}}$<br><br>Option (b).
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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