Q.A manufacturer produces switches and experiences that 2 percent switches are defective. The probability that in a box of 50 switches, there are atmost two defective is ______.

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).

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

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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 $x=0$

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

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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 $m$.

   $\text{Var}(X)=m=2$ …

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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 $P(X=1)$

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

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

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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. …

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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 $x=10,\ \lambda=5$

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

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

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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 $e^{-3}$). …

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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 …

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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 $m=m^2/2$. …

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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. …

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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 $SD=\sqrt{\lambda}=\sqrt{9}=3.$ …

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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. …

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