Two fair dice are thrown. The numbers on them | vedclass

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(B) The system has a unique solution if the determinant $D 
eq 0$.$D = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & \lambda \end{vmatrix} = 1(2\

Vedclass · Accepted Answer

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

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(B) The system has a unique solution if the determinant $D 
eq 0$.$D = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & \lambda \end{vmatrix} = 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2) = 2\lambda - 9 - \lambda + 3 + 1 = \lambda - 5$.For a unique solution,$D 
eq 0 \Rightarrow \lambda 
eq 5$.Since $\lambda$ is the outcome of a die,$\lambda \in \{1, 2, 3, 4, 5, 6\}$. The condition $\lambda 
eq 5$ is satisfied by $5$ outcomes out of $6$. Thus,$p = \frac{5}{6}$.For no solution,$D = 0 \Rightarrow \lambda = 5$. We check the consistency using $D_1, D_2, D_3$.$D_1 = \begin{vmatrix} 5 & 1 & 1 \ \mu & 2 & 3 \ 1 & 3 & 5 \end{vmatrix} = 5(10-9) - 1(5\mu-3) + 1(3\mu-2) = 5 - 5\mu + 3 + 3\mu - 2 = 6 - 2\mu$.For no solution,$D=0$ and at least one of $D_1, D_2, D_3 
eq 0$. If $\lambda=5$,$D_1 = 6-2\mu$. $D_1 
eq 0 \Rightarrow \mu 
eq 3$.Since $\mu$ can be any of the $6$ outcomes,the probability that $\mu 
eq 3$ is $\frac{5}{6}$.Thus,$q = P(\lambda=5) 	imes P(\mu 
eq 3) = \frac{1}{6} 	imes \frac{5}{6} = \frac{5}{36}$.

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:

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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: