Q.In a large batch of 100 items, the probability that any one item is defective is 0.02. Using the Poisson approximation to the binomial distribution, find the probability that exactly 1 item in the batch is defective. (Use e^{-2} ≈ 0.1353.)

Showing the 12 most recent of 16 on this concept.

1. CA Foundation 2026Set jan-20261 markMCQ
   Q.If X is a Poisson variate such that $P(X=1)=0.3$, $P(X=2)=0.2$, then $P(X=0)=$ (A) $e^{\frac{4}{3}}$ (B) $e^{\frac{-1}{3}}$ (C) $e^{\frac{-4}{3}}$ (D) $e^{\frac{-2}{3}}$

   ›Reveal solutionSolution

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

   Step 1 — form the ratio.

   $\frac{P(X=2)}{P(X=1)} = \frac{e^{-\lambda}\lambda^2/2!}{e^{-\lambda}\lambda} = \frac{\lambda}{2}.$

   Step 2 — substitute the given probabilities.

   $\frac{\lambda}{2} = \frac{0.2}{0.3} = \frac{2}{3} \Rightarrow \lambda = \frac{4}{3}.$

   Step 3 — compute P(X=0).

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

   Watch out

   The exponent stays negative: $P(X=0) = e^{-\lambda} = e^{-4/3}$, not the positive $e^{4/3}$ (option A).

   Tip

   Taking the ratio of consecutive Poisson probabilities cancels $e^{-\lambda}$ and simplifies the factorials, leaving just $\lambda / k$.

   ✓Final answer

   (C) e^(−4/3)

2. CA Foundation 2026Set may-20261 markMCQ
   Q.If the standard deviation of a Poisson distribution is 3, then $P(X = 0)$ is ________. (A) $e^{-6}$ (B) $e^{-3}$ (C) $e^{-9}$ (D) $e^{-1}$

   ›Reveal solutionSolution

   Poisson: $\lambda=\sigma^2=9$, so $P(X=0)=e^{-\lambda}=e^{-9}$.

   Step 1 — Get the parameter $\lambda$

   For a Poisson variable the mean and variance are both $\lambda$, hence the standard deviation is $\sqrt{\lambda}$:

   $\sigma=\sqrt{\lambda}=3 \;\Rightarrow\; \lambda=9$

   Step 2 — Apply the Poisson probability formula

   $P(X=x)=\dfrac{e^{-\lambda}\lambda^{x}}{x!}$

   Step 3 — Put $x=0$

   $P(X=0)=\dfrac{e^{-9}\,9^{0}}{0!}=e^{-9}$

   Watch out

   The standard deviation is 3, but the parameter is the variance $\lambda=9$, not 3 — using $\lambda=3$ wrongly gives $e^{-3}$ (option B).

   Tip

   For Poisson, always square the SD to recover $\lambda$ before using any probability formula.

   ✓Final answer

   (C) $e^{-9}$

3. CA Foundation 2025Set jan-20251 markMCQ
   Q.If X is a Poisson variable such that $P(X=1) = P(X=2)$ then the variance is (A) 2 (B) 1 (C) $\sqrt{2}$ (D) 3

   ›Reveal solutionSolution

   $P(X=1)=P(X=2)\Rightarrow m=2$, and for Poisson variance $=$ mean $=2$.

   Step 1 — Set the two probabilities equal

   $P(X=1)=e^{-m}m,\qquad P(X=2)=\frac{e^{-m}m^{2}}{2}$

   $e^{-m}m=\frac{e^{-m}m^{2}}{2}$

   Step 2 — Solve for the parameter $m$

   Cancel $e^{-m}m$ (with $m\ne0$):

   $1=\frac{m}{2}\ \Rightarrow\ m=2$

   Step 3 — Read off the variance

   For a Poisson distribution the mean and variance are both equal to $m$.

   $\text{Var}(X)=m=2$

   Why the other options are wrong: 1 is $m$ if you mistakenly cancel to $m=1$; $\sqrt2$ is the standard deviation confused with variance; 3 has no basis.

   Watch out

   The question asks for variance, not standard deviation — variance $=2$, SD $=\sqrt2$ (the trap option C).

   Tip

   Poisson signature: mean $=$ variance $=m$. Any 'find the variance' Poisson question reduces to finding $m$.

   ✓Final answer

   (A) 2

4. CA Foundation 2025Set jan-20251 markMCQ
   Q.If 3 percent of ceramic cup manufactured by a company are known to be defective. What is the probability that a sample of 100 cups are taken from the production process, of that company would contain exactly one defective cup? (A) 0.15 (B) 0.03 (C) 0.09 (D) 0.30

   ›Reveal solutionSolution

   Poisson with $\lambda=np=3$: $P(1)=e^{-3}\cdot3\approx0.15$.

   Step 1 — Set up the Poisson approximation

   Here $n=100$ is large and $p=0.03$ is small, so binomial $\approx$ Poisson with

   $\lambda=np=100\times0.03=3$

   Step 2 — Compute $P(X=1)$

   $P(X=r)=\frac{e^{-\lambda}\lambda^{r}}{r!}$

   $P(X=1)=e^{-3}\cdot\frac{3^{1}}{1!}=3e^{-3}=3\times0.0498=0.1494\approx0.15$

   Why the other options are wrong: $0.03$ is just $p$; $0.09$ and $0.30$ don't follow from the Poisson formula.

   Watch out

   Use $\lambda=np$, not $p$, as the Poisson parameter — plugging $0.03$ directly is the classic error.

   Tip

   $e^{-3}\approx0.0498$ is worth memorising for CA Poisson MCQs.

   ✓Final answer

   (A) 0.15

5. CA Foundation 2025Set may-20251 markMCQ
   Q.Poisson probability distribution is appropriately applied in (A) The height of students in the university. (B) The distribution of passing of students in university examinations. (C) Tossing of a coin hundred times. (D) Number of deaths by a rare disease.

   ›Reveal solutionSolution

   Poisson applies to counts of rare events with small probability over many trials — the deaths-from-a-rare-disease case.

   Step 1 — What Poisson describes

   Important

   The Poisson distribution counts how often a rare event happens in a given time/space, when the number of trials $n$ is large and the success probability $p$ is very small (with $\lambda=np$ moderate).

   Step 2 — Test each option

   • (A) Height of students — a CONTINUOUS variable → Normal distribution, not Poisson.
   • (B) Passing of students — a proportion of successes with moderate $p$ → Binomial.
   • (C) Tossing a coin 100 times — fixed $n$, $p=0.5$ (not small) → Binomial.
   • (D) Deaths by a rare disease — a rare count over a population → Poisson.

   Watch out

   Poisson needs the event to be RARE (small $p$). A coin toss with $p=0.5$ is not rare, so (C) is binomial, not Poisson.

   Tip

   Cue words for Poisson: "rare", "number of", "per unit time/area", "accidents/defects/deaths".

   ✓Final answer

   (D) Number of deaths by a rare disease.

6. CA Foundation 2025Set may-20251 markMCQ
   Q.If 5% of the families in large population city do not use gas as a fuel, what will be the probability of selecting 10 families in a random sample of 100 families who do not use gas as a fuel ? [Given that $e^{-5} = 0.0067$] (A) 0.038 (B) Zero (C) 0.018 (D) 0.048

   ›Reveal solutionSolution

   Poisson with $\lambda=np=5$: $P(X=10)=\dfrac{e^{-5}5^{10}}{10!}\approx0.018$.

   Step 1 — Find the Poisson parameter

   $5\%$ of families do not use gas, so in a sample of 100:

   $\lambda=np=100\times0.05=5$

   Since $p$ is small and $n$ large, use the Poisson approximation.

   Step 2 — Apply the Poisson formula

   $P(X=x)=\dfrac{e^{-\lambda}\lambda^{x}}{x!}$

   Step 3 — Substitute $x=10,\ \lambda=5$

   $P(X=10)=\dfrac{e^{-5}\times5^{10}}{10!}=\dfrac{0.0067\times9765625}{3628800}$

   $P(X=10)=\dfrac{65429.7}{3628800}\approx0.018$

   Why the other options are wrong: (A) 0.038 and (D) 0.048 mis-evaluate the factorial/power; (B) Zero wrongly assumes 10 is impossible.

   Watch out

   The mean is $\lambda=np=5$, NOT $p=0.05$ — a very common slip that would make the whole computation wrong.

   Tip

   Large $n$ + small $p$ = Poisson; compute $\lambda=np$ first, then plug into $\dfrac{e^{-\lambda}\lambda^{x}}{x!}$.

   ✓Final answer

   (C) 0.018

7. CA Foundation 2025Set sep-20251 markMCQ
   Q.An emergency room receives an average of 3 patients per hour. What is the probability that exactly 2 patients arrive in an hour ? (Given : $e^0 = 1, e^{-1} = 0.367, e^{-2} = 0.135, e^{-3} = 0.049, e^{-4} = 0.018, e^{-5} = 0.0067$) (A) 0.22 (B) 0.3 (C) 0.27 (D) 0.25

   ›Reveal solutionSolution

   Poisson with λ = 3: P(X=2) = e⁻³·3²/2! = 0.049·9/2 ≈ 0.22.

   Step 1 — Identify the model and parameter

   Random arrivals at a constant average rate → Poisson, with $\lambda=3$ patients per hour.

   $P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!}$

   Step 2 — Substitute x = 2

   $P(X=2)=\frac{e^{-3}\cdot 3^{2}}{2!}=\frac{0.049\times 9}{2}=\frac{0.441}{2}=0.2205$

   Rounded, $P(X=2)\approx0.22$.

   Why the other options are wrong: 0.30, 0.27, 0.25 result from using the wrong power of λ or the wrong value of $e^{-3}$ (e.g. mixing up $e^{-2}$ with $e^{-3}$).

   Watch out

   Use $e^{-\lambda}=e^{-3}$, not $e^{-2}$ — the exponent is the mean λ, not the value x you are computing the probability for.

   Tip

   Keep λ, x and the exponent straight: exponent is always $-\lambda$; the power of λ and the factorial both use x.

   ✓Final answer

   (A) 0.22

8. CA Foundation 2024Set sep-20241 markMCQ
   Q.The number of accidents in a year attributed to taxi drivers in a locality follows Poisson distribution with average 2. Out of 500 taxi drivers of that area, what is the number of drivers with at least 3 accidents in a year ? (Given that e = 2.718) (A) 162 (B) 180 (C) 201 (D) 190

   ›Reveal solutionSolution

   $P(X\ge 3)=1-P(0)-P(1)-P(2)=0.3233$; expected drivers $=500\times0.3233\approx 162$.

   Step 1 — Poisson probabilities with mean $m=2$

   $P(X=x) = \frac{e^{-m} m^x}{x!}, \quad e^{-2} = \frac{1}{2.718^2} = 0.1353$

   $P(0) = e^{-2} = 0.1353$

   $P(1) = 2e^{-2} = 0.2707$

   $P(2) = \frac{2^2}{2!}e^{-2} = 2e^{-2} = 0.2707$

   Step 2 — Probability of at least 3 accidents

   $P(X\ge 3) = 1 - (0.1353 + 0.2707 + 0.2707) = 1 - 0.6767 = 0.3233$

   Step 3 — Expected number of such drivers

   $500 \times 0.3233 = 161.6 \approx 162$

   Why the other options are wrong

   • (B) 180, (C) 201, (D) 190 come from arithmetic slips in $e^{-2}$ or from computing $P(X\ge 2)$ / $P(X>3)$ instead of $P(X\ge 3)$.

   Watch out

   "At least 3" means $X\ge 3$, so subtract $P(0),P(1),P(2)$ — three terms. Subtracting only $P(0),P(1)$ (i.e. "at least 2") over-counts.

   Tip

   Compute $e^{-2}$ once and reuse it; $P(1)=P(2)$ here because $m=2$, a handy check.

   ✓Final answer

   (A) 162

9. CA Foundation 2024Set sep-20241 markMCQ
   Q.If a random variable X follows Poisson distribution such that $P(X = 1) = P(X = 2)$, then the mean of the distribution is : (A) 2 (B) 1 (C) 0 (D) 1/2

   ›Reveal solutionSolution

   $P(1)=P(2)\Rightarrow m=\dfrac{m^2}{2}\Rightarrow m=2$.

   Step 1 — Write the two probabilities

   $P(X=1) = \frac{e^{-m}m^1}{1!} = m e^{-m}$

   $P(X=2) = \frac{e^{-m}m^2}{2!} = \frac{m^2}{2} e^{-m}$

   Step 2 — Equate and cancel $e^{-m}$

   $m e^{-m} = \frac{m^2}{2} e^{-m} \;\Rightarrow\; m = \frac{m^2}{2}$

   Step 3 — Solve

   $2m = m^2 \;\Rightarrow\; m^2 - 2m = 0 \;\Rightarrow\; m(m-2)=0$

   $m=0$ is rejected (a Poisson mean must be positive), so $m = 2$. For a Poisson distribution the mean equals $m$.

   Why the other options are wrong

   • (B) 1, (D) 1/2 do not satisfy $m=m^2/2$.
   • (C) 0 is the rejected root (a degenerate, non-random case).

   Watch out

   Discard the root $m=0$: a Poisson parameter must be strictly positive, so the meaningful solution is $m=2$.

   Tip

   Cancelling $e^{-m}$ (never zero) turns the condition into a simple quadratic in $m$.

   ✓Final answer

   (A) 2

10. CA Foundation 2023Set jun-20231 markMCQ
    Q.Between 9 AM and 10 AM, the average number of phone calls per minute coming into the switchboard of a company is 4. Find the probability that during one particular minute, there will be either 2 phone calls or no phone calls (given $e^{-4}=0.018316$). (A) 0.156 (B) 0.165 (C) 0.149 (D) 0.194

    ›Reveal solutionSolution

    P(0) + P(2) = 0.018316 + 0.146528 ≈ 0.165 for a Poisson mean of 4.

    Step 1 — Model as Poisson

    Calls per minute follow Poisson with $\lambda = 4$, so $P(X=x) = \dfrac{e^{-\lambda}\lambda^{x}}{x!}$.

    Step 2 — Compute P(0) and P(2)

    $P(0) = e^{-4} = 0.018316$

    $P(2) = \frac{e^{-4}\,4^{2}}{2!} = 0.018316 \times \frac{16}{2} = 0.018316 \times 8 = 0.146528$

    Step 3 — Add the mutually exclusive cases

    $P(0 \text{ or } 2) = 0.018316 + 0.146528 = 0.164844 \approx 0.165$

    Watch out

    "Either 2 or none" means add P(2) and P(0); do not multiply them — these are mutually exclusive counts, not independent events.

    Tip

    Factor out $e^{-4}$: $P(0)+P(2) = e^{-4}\left(1 + \frac{16}{2}\right) = 9e^{-4} = 9 \times 0.018316 \approx 0.165$.

    ✓Final answer

    (B) 0.165

    Note

    This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.

11. CA Foundation 2023Set jun-20231 markMCQ
    Q.If a Poisson distribution is such that $P(X=2)=\dfrac{1}{3}P(X=3)$, then the standard deviation of the distribution is : (A) $\sqrt{3}$ (B) 3 (C) 2 (D) 1

    ›Reveal solutionSolution

    Solving P(X=2)=⅓P(X=3) gives λ=9, so SD = √λ = 3.

    Step 1 — Write the Poisson probabilities

    $P(X=2)=\frac{e^{-\lambda}\lambda^{2}}{2!},\qquad P(X=3)=\frac{e^{-\lambda}\lambda^{3}}{3!}.$

    Step 2 — Apply the given relation

    $\frac{\lambda^{2}}{2}=\frac{1}{3}\cdot\frac{\lambda^{3}}{6}=\frac{\lambda^{3}}{18}.$

    Divide by λ²: $\dfrac{1}{2}=\dfrac{\lambda}{18}\Rightarrow \lambda = 9.$

    Step 3 — Standard deviation

    For Poisson, variance = mean = λ, so $SD=\sqrt{\lambda}=\sqrt{9}=3.$

    Watch out

    A common slip is to stop at λ=9 and report it (option is not there) or to give the variance instead of the SD. The SD is √λ, not λ.

    Tip

    Remember Poisson's signature: mean = variance = λ, hence SD = √λ.

    ✓Final answer

    (B) 3

    Note

    This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.

12. CA Foundation 2022Set dec-20221 markMCQ
    Q.If a Poisson distribution is such that $P(X=2) = P(X=3)$ then the variance of the distribution is (A) $\sqrt{3}$ (B) 3 (C) 6 (D) 9

    ›Reveal solutionSolution

    P(X=2)=P(X=3) ⇒ λ = 3, and variance = λ = 3.

    Step 1 — Equate the probabilities

    $\frac{e^{-\lambda}\lambda^2}{2!} = \frac{e^{-\lambda}\lambda^3}{3!}$

    Step 2 — Simplify

    $\frac{1}{2} = \frac{\lambda}{6} \Rightarrow \lambda = 3$

    Step 3 — Variance

    For Poisson, mean = variance = λ, so variance = 3.

    Watch out

    (A) √3 confuses variance with standard deviation; for Poisson the variance itself is λ.

    Tip

    Cancel e⁻λ and one power of λ immediately to isolate λ.

    ✓Final answer

    (B) 3

    Note

    This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.