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Let $X_1, X_2, X_3, X_4$ be i.i.d random variables each assuming the values $1$ and $-1$ with probabilities $1/2$. then the probability that the matrix

\begin{pmatrix} X_1 & X_2\\ X_3 & X_4 \end{pmatrix}

is nonsingular?

We have to find out probability of $X_1X_4-X_2X_3 \neq 0$, but how to do?

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    $\begingroup$ Worst comes to worst, there are $16$ cases to check. $\endgroup$
    – user228113
    Commented Jan 25, 2018 at 6:47
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    $\begingroup$ What could be easier than checking the 16 cases? Draw a small picture that shows that 4 possibilities that columns can take. The answer is easy to see from there. $\endgroup$
    – copper.hat
    Commented Jan 25, 2018 at 6:55

2 Answers 2

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Either $X_1X_4=X_2X_3=1$ ($4$ ways) or $X_1X_4=X_2X_3=-1$ ($4$ ways), so $\frac 8{2^4}=\frac 12$ probability...

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It is perhaps (geometrically) easier to check if ${X_1 \over X_2} = {X_3 \over X_4}$. Either side takes values $\pm 1$ with equal probability.

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