Let N denote the number that turns up when a fair die is rolled. If the probability that the system of equations
$x + y + z = 1$
$2x + \mathrm{N}y + 2z = 2$
$3x + 3y + \mathrm{N}z = 3$
has unique solution is ${k \over 6}$, then the sum of value of k and all possible values of N is :
Solution
For unique solution $\Delta \neq 0$
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i.e. $\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N\end{array}\right| \neq 0$
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$\Rightarrow\left(N^{2}-6\right)-(2 N-6)+(6-3 N) \neq 0$
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$\Rightarrow N^{2}-5 N+6 \neq 0$
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$\therefore N \neq 2$ and $N \neq 3$
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$\therefore $ Probability of not getting 2 or 3 in a throw of dice $=\frac{2}{3}$
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As given $\frac{2}{3}=\frac{k}{6} \Rightarrow k=4$
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$\therefore$ Required value $=1+4+5+6+4=20$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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