ProbabilityHard

Question

Let n be the number obtained on rolling a fair die. If the probability that the system

$${x - ny + z = 6 }{x + (n - 2)y + (n + 1)z = 8}$$

$$(n - 1)y + z = 1$$

Has a unique solution is $\frac{k}{6}$, then the sum of k and all possible values of n is :

Options

A.21
B.24
C.20
D.22

Solution

$x - ny + z = 6$

$${x + (n - 2)y + (n + 1)z = 8 }{(n - 1)y + z = 1 }{\left| \begin{matrix} 1 & - n & 1 \\ 1 & (n - 2) & n + 1 \\ 0 & n - 1 & 1 \end{matrix} \right| \neq 0 }{\Rightarrow n^{2} - 3n + 2 \neq 0 }{n \neq 1,2 }$$for unique solution $n = 3,4,5,6$

Now

P (probability when system of equations has unique solution) $= \frac{4}{6}$

So $k = 4$

Now required sum $= 4 + (3 + 4 + 5 + 6) = 22$

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