CA Foundation 2026 · Paper 3 · Quantitative AptitudeQ95 · 1 mark↻ Appears in 6 of 6 yearsOfficial key verified
If X is a Poisson variate such that P(X=1)=0.3P(X=1)=0.3, P(X=2)=0.2P(X=2)=0.2, then P(X=0)=P(X=0)=
Figure for question 95
Single correct — pick one, then checkICAI format
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Poisson: P(X=k)=e−λλkk!P(X=k) = \dfrac{e^{-\lambda}\lambda^k}{k!}; the ratio of consecutive probabilities isolates λ\lambda.

Step 1 — form the ratio.

P(X=2)P(X=1)=e−λλ2/2!e−λλ=λ2.\frac{P(X=2)}{P(X=1)} = \frac{e^{-\lambda}\lambda^2/2!}{e^{-\lambda}\lambda} = \frac{\lambda}{2}.

Step 2 — substitute the given probabilities.

λ2=0.20.3=23⇒λ=43.\frac{\lambda}{2} = \frac{0.2}{0.3} = \frac{2}{3} \Rightarrow \lambda = \frac{4}{3}.

Step 3 — compute P(X=0).

P(X=0)=e−λ=e−4/3.P(X=0) = e^{-\lambda} = e^{-4/3}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.

← Back to the Paper 3 · Quantitative Aptitude 2026 paper