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Part CCSIR NET December 2024a-free-variable-needs-consistency-too

A free variable needs consistency too

Let Z1Z_{1} and Z2Z_{2} be two independent discrete random variables such that Z1Z_{1} follows binomial distribution with parameters n = 2 and p = 1/2, and Z2Z_{2} follows Poisson distribution with mean 1. Consider the following system of equations with three variables x1,x2x_{1}, x_{2} and x3:x1−2x2+x3=1;2x1−5x2+2x3=2;x1+2x2+Z1x3=Z2x_{3}: x_{1} - 2x_{2} + x_{3} = 1; 2x_{1} - 5x_{2} + 2x_{3} = 2; x_{1} + 2x_{2} + Z_{1}x_{3} = Z_{2}. Then, the probability that the given system of equations has infinite number of solutions equals

  1. A.e⁻1^{1}
  2. B.e⁻1/2^{1}/2✓ Correct
  3. C.e⁻1/4^{1}/4
  4. D.e⁻1/12^{1}/12

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests existence vs uniqueness.

See pricing

50 are analysed free — try those first.

The trap it tests

Existence vs uniqueness

A theorem giving one was read as giving both.

Drill statements like this

Related counterexample: The maximum of independent exponentials is exponential

More on this topic

The chapter behind this: The standard distributions — free to read

From Probability › Standard discrete and continuous distributions

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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