ISI-2014-11

Question

Let $X_1,X_2,X_3,X_4$ be i.i.d. random variables each assuming the value $1$ and $-1$ with probability $\dfrac{1}{2}$ each. Then, the probability that the matrix $\begin{pmatrix}X_1 &X_2\\ X_3 &X_4\end{pmatrix}$ is nonsingular equals

• A. $1/2$
• B. $3/8$
• C. $5/8$
• D. $1/4$

Answer 1

For matrix to be nonsingular any of these cases can occur$:$

 $X_{1}=1,X_{2}=1,X_{3}=-1,X_{4}=1$                       $ X1=-1,X2=1,X3=1,X4=1$

 $X_{1}=1,X_{2}=-1,X_{3}=1,X_{4}=1$                       $ X_{1}=-1,X_{2}=-1,X_{3}=-1,X_{4}=1$

$X_{1}=-1,X_{2}=1,X_{3}=-1,X_{4}=-1$               $ X_{1}=1,X_{2}=-1,X_{3}=-1,X_{4}=-1$

$X_{1}=-1,X_{2}=-1,X_{3}=1,X_{4}=-1 $             $X_{1}=1,X_{2}=-1,X_{3}=-1,X_{4}=-1$

each case has probability $=(\frac{1}{2})^{4} = \frac{1}{16}$

so required probability $=8\times\frac{1}{16}$ = $\frac{1}{2}$

So option $(A)$  is correct

Comments

yes, i think you're right, even im getting (1/2)

Answer 2

$X1\times X4-X2\times X3=0$

$X1\times X2=X2\times X3$

So$,$Number of value satisfy equation is$:$

$\text{1) All 1}$

$\text{2) All -1}$

$\text{3) Even number of 1 and -1 = $\frac{4!}{2!*2!}=6$}$

Total cases are $4^{n}$ where $n=2$ as $-1$ and $1$ and $4$ because of $4$ variable

Probability will be $\frac{1+1+6}{16}=\frac{1}{2}$

Answer 3

I think answer should be 8/16 total possible sample space outcomes =16

fav outcome 8 so 8/16 as answer here

Comments

how are possible sample space outcomes =8 ? isn't it 16 ?

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yes it would be 16