CUET UG Applied Mathematics Booster Test 3

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CUET UG Applied Mathematics Booster Test 3 - Poisson Distribution

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 3 - Poisson Distribution

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QUESTION 1 OF 20

The mortality rate for a certain disease is 0.007. In a group of 400 people, the occurrence of deaths is modeled via Poisson. What is the value of the expected mean parameter λ?

Answer Review

1 The mortality rate for a certain disease is 0.007. In a group of 400 people, the occurrence of deaths is modeled via Poisson. What is the value of the expected mean parameter λ?

A 0.007

B 2.800

C 4.000

D 28.000

💡 Short Explanation

�� Poisson mean λ = np �� n = 400, p = 0.007 �� λ = 2.8

✅ Correct Answer Explanation

For Poisson approximation of binomial events: λ=np Substituting values: λ = 400 × 0.007 = 2.8 So expected number of deaths = 2.8.

❓ Why Other Options Are Incorrect

�� Option A → Only probability, not expected count
�� Option C → Overestimation
�� Option D → Incorrect scaling
Used:
Substitution

Application: Direct use of λ = np

Final Logic: Only 400 × 0.007 fits Poisson mean

🧠 Memory Trick

"Poisson mean = number × probability"

📚 Class XII – NCERT Mathematics Part II – Chapter: Probability – Topic: Poisson Approximation –

2 Match the specific Poisson probability calculations for the river flood model (λ = 2) to their simplified analytical expressions.

List I	List II
(A) P(X = 0)	(I) (4/3) × e^-2
(B) P(X = 1)	(II) 2 × e^-2
(C) P(X = 2)	(III) e^-2
(D) P(X = 3)	(IV) 2 × e^-2

A (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

B (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

C (A)-(III), (B)-(I), (C)-(II), (D)-(IV)

D (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

💡 Short Explanation

�� Apply Poisson formula �� Substitute λ = 2 �� Match k-values

✅ Correct Answer Explanation

Poisson formula: P(X=k)=λ^ke^(-λ)/k! For λ = 2: • P(X=0) = e⁻² → (III) • P(X=1) = 2e⁻² → (II) • P(X=2) = (2² e⁻²)/2! → matches (IV) • P(X=3) = (4/3)e⁻² → matches (I) Hence correct mapping is option A.

❓ Why Other Options Are Incorrect

�� Option B → swaps k=0 and k=2 expressions
�� Option C → incorrect pairing logic
�� Option D → inconsistent factorial usage
Used:
Option Grouping

Application: Match each k separately using formula

Final Logic: Only option A preserves correct structure

🧠 Memory Trick

"0 → pure exponential, 1 → λ times exponential"

📚 Class XII – NCERT Mathematics Part II – Chapter: Probability – Topic: Poisson Distribution –

3 Which of the following distributions accurately transition into or share properties with the Poisson model given infinite independent trials?
(A) Binomial distribution as n → ∞ and p → 0.
(B) A discrete distribution where events occur randomly over continuous time.
(C) Standard Normal Distribution where Z-score is always 0.
(D) Models predicting discrete rare event occurrences.

A A,B,D

B A,B

C A,C,D

D A,B,C,D

💡 Short Explanation

�� Poisson limit of binomial �� Independent rare events �� Continuous-time modeling

✅ Correct Answer Explanation

Poisson distribution arises as a limit of binomial when: n→∞, p→0, np=λ Thus: • Binomial convergence is valid • Random independent event processes fit • Rare event models apply Normal distribution (option C) is unrelated to Poisson framework.

❓ Why Other Options Are Incorrect

�� Option C → includes normal distribution incorrectly
�� Option B → missing valid case
�� Option D → incorrect inclusion of unrelated distribution
Used:
Elimination

Application: Keep only Poisson-compatible models

Final Logic: Remove normal distribution case

🧠 Memory Trick

"Poisson = rare + limit of binomial"

📚 Class XII – NCERT Mathematics Part II – Chapter: Probability – Topic: Poisson Approximation –

4 Identify the INCORRECT analytical property regarding the Poisson parameter λ.

A λ = n × p

B Standard deviation = √λ

C λ must be negative

D Variance = λ

💡 Short Explanation

�� λ is mean rate �� Always non-negative �� Defines distribution

✅ Correct Answer Explanation

In Poisson distribution: • λ represents expected mean occurrences • λ > 0 always • Variance = λ • Standard deviation = √λ So λ being negative is impossible.

❓ Why Other Options Are Incorrect

�� Option A → valid approximation
�� Option B → correct property
�� Option D → fundamental identity
Used:
Extreme Word Filter

Application: Check validity of parameter sign

Final Logic: Only negative λ violates definition

🧠 Memory Trick

"λ can be zero or positive, never negative"

📚 Class XII – NCERT Mathematics Part II – Chapter: Probability – Topic: Poisson Properties –

5 3% of electric bulbs are defective. In a sample of 100 bulbs modeled by Poisson (λ = 3), what is P(X=0)?

A 0.03

B 0.05

C 0.95

D 0.15

💡 Short Explanation

�� k = 0 case �� Poisson formula �� e⁻³ value

✅ Correct Answer Explanation

Poisson probability: P(X=0)=e^(-λ) So for λ = 3: P(X=0) = e⁻³ ≈ 0.05

❓ Why Other Options Are Incorrect

�� Option A → incorrect probability
�� Option C → complement confusion
�� Option D → overestimated value
Used:
Substitution

Application: direct k=0 simplification

Final Logic: only e⁻λ fits

🧠 Memory Trick

"Zero event = pure exponential"

📚 Class XII – NCERT Mathematics Part II – Chapter: Probability – Topic: Poisson Formula –

6 Within a computer disk quality constraint region, errors follow a Poisson distribution with λ = 0.2. What is the probability of exactly one error?

A 0.2 * e⁻0.2

B e⁻0.2

C (0.04 * e⁻0.2) / 2

D 1 - e⁻0.2

💡 Short Explanation

�� k = 1 case �� Apply Poisson formula �� λ¹ term appears

✅ Correct Answer Explanation

Poisson formula: P(X=1)=λe^(-λ) Substitute λ = 0.2: P(X=1) = 0.2 × e⁻0.2

❓ Why Other Options Are Incorrect

�� Option B → k=0 case
�� Option C → wrong factorial structure
�� Option D → complement event
Used:
Substitution

Application: k=1 direct formula use

Final Logic: only λe⁻λ valid

🧠 Memory Trick

"Single event = lambda × exponential"

📚 Class XII – NCERT Mathematics Part II – Chapter: Probability – Topic: Poisson Distribution –

7 (Expected Mean Index) In the airport passenger problem, the overall hourly expected index is λ = 30. If we split the hour into a 10-minute segment and a 50-minute segment, the sum of their individual λ parameters equals:

A 6 + 24

B 5 + 25

C 10 + 20

D 15 + 15

💡 Short Explanation

The Poisson mean (λ) is directly proportional to the length of the time interval. A 10-minute interval is ( \frac{1}{6} ) of an hour. The remaining 50 minutes account for ( \frac{5}{6} ) of the hourly mean.

✅ Correct Answer Explanation

For a Poisson process, [ \lambda_{\text{new}}=\lambda_{\text{hour}}\times\frac{\text{New Time}}{\text{Original Time}} ] Given, Hourly mean (=30) For 10 minutes: [ 30\times\frac{10}{60}=5 ] For 50 minutes: [ 30\times\frac{50}{60}=25 ] Thus, [ 5+25=30 ] which equals the original hourly expectation. Therefore, Option B is correct. Option A (6 + 24): Total is correct but the interval values are not proportional. Option B (5 + 25): Correct proportional division. Option C (10 + 20): Incorrect scaling. Option D (15 + 15): Represents equal 30-minute intervals, not 10 and 50 minutes.

❓ Why Other Options Are Incorrect

Option A) 6 + 24 → Incorrect λ values for the given intervals.
Option C) 10 + 20 → Does not match the 10:50 time ratio.
Option D) 15 + 15 → Corresponds to two equal half-hour intervals.

🎯 Strategy

Used

Substitution

Application: Convert each time interval into a fraction of one hour and multiply by the hourly λ.

Final Logic: Since (30\times\frac{10}{60}=5) and (30\times\frac{50}{60}=25), Option B is correct.

🧠 Memory Trick

"New λ = Rate × Time Fraction."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Poisson Distribution (Parameter λ)

8 If the moving average (E(X)) of a Poisson variable is 2.8 (mortality rate in 400 people), what is the formula to calculate the standard deviation of deaths?

A 2.8

B (2.8)²

C √2.8

D 2.8 / 2

💡 Short Explanation

For a Poisson distribution, mean equals variance. Standard deviation is the square root of the variance. Therefore, SD = √2.8.

✅ Correct Answer Explanation

For a Poisson distribution, [ E(X)=\lambda,\qquad \text{Variance}=\lambda ] Hence, [ \text{Standard Deviation}=\sqrt{\lambda} ] Given, [ \lambda=2.8 ] Therefore, [ \boxed{\text{SD}=\sqrt{2.8}} ] Thus, Option C is correct. Option A: Represents the mean, not the standard deviation. Option B: Squares the mean incorrectly. Option C: Correct formula. Option D: Not a statistical formula.

❓ Why Other Options Are Incorrect

Option A) 2.8 → Mean (and variance), not standard deviation.
Option B) (2.8)² → Incorrect computation.
Option D) 2.8 / 2 → No such formula exists.

🎯 Strategy

Used

Formula Recall

Application: Recall that Poisson variance equals λ and SD equals √λ.

Final Logic: Since SD = √λ, the answer is √2.8, making Option C correct.

🧠 Memory Trick

"Poisson: Mean = Variance = λ, SD = √λ."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Poisson Distribution

9 The sum of all probabilities in a Poisson distribution must equal 1. This condition relies on the infinite Maclaurin series expansion of which mathematical term?

A e^λ

B e^-λ

C λ^k

D ln(λ)

💡 Short Explanation

The Poisson probability mass function contains the factor (e^{-\lambda}). The infinite series for (e^\lambda) normalizes the distribution. Their product makes the total probability equal to 1.

✅ Correct Answer Explanation

The Poisson distribution is [ P(X=k)=\frac{\lambda^k e^{-\lambda}}{k!} ] Summing over all values, [ \sum_{k=0}^{\infty}\frac{\lambda^k e^{-\lambda}}{k!} =e^{-\lambda}\sum_{k=0}^{\infty}\frac{\lambda^k}{k!} ] Using the Maclaurin expansion, [ e^\lambda=\sum_{k=0}^{\infty}\frac{\lambda^k}{k!} ] Therefore, [ e^{-\lambda}\times e^\lambda=1 ] Hence, Option A is correct. Option A: Correct series used in normalization. Option B: Appears in the PMF but is not expanded in the normalization step. Option C: Only part of the numerator. Option D: Unrelated.

❓ Why Other Options Are Incorrect

Option B) e^-λ → Multiplicative factor, not the expanded series.
Option C) λ^k → Individual term, not the complete series.
Option D) ln(λ) → Not involved in the normalization.

🎯 Strategy

Used

Conceptual Recall

Application: Recall the Maclaurin expansion used to prove the total probability equals one.

Final Logic: The expansion of (e^\lambda) ensures normalization, making Option A correct.

🧠 Memory Trick

"Poisson Normalization: (e^{-\lambda}) × (e^\lambda) = 1."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Poisson Distribution (Normalization)

10 For a large vector of trials modeling customer arrivals (n = 3600 seconds, p = 1/120), the Poisson distribution establishes that the probability of MORE THAN ONE arrival during a single second interval infinitesimally approaches:

A 1

B 0

C p

D λ

💡 Short Explanation

A Poisson process assumes events occur one at a time in very small intervals. The probability of two or more events in an infinitesimal interval is negligible. Hence, it approaches zero.

✅ Correct Answer Explanation

One assumption of a Poisson process is: The probability of one event occurring in a very small interval is proportional to the interval length. The probability of more than one event in that tiny interval is negligible. Therefore, [ P(\text{more than one event})\rightarrow0 ] Thus, Option B is correct. Option A: Impossible for an infinitesimal interval. Option B: Correct limiting property. Option C: Not the required probability. Option D: λ is a parameter, not a probability.

❓ Why Other Options Are Incorrect

Option A) 1 → Contradicts Poisson assumptions.
Option C) p → Represents success probability, not this limit.
Option D) λ → Mean parameter, not a probability.

🎯 Strategy

Used

Conceptual Recall

Application: Recall the assumptions underlying a Poisson process.

Final Logic: In an infinitesimal interval, the probability of more than one arrival tends to 0, making Option B correct.

🧠 Memory Trick

"Tiny Interval → Two Events Impossible."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Poisson Process Assumptions

11 An ice-cream parlour receives 4 customers per minute. Modeling a 4-minute area block yields λ = 16. What is the probability of receiving exactly 0 customers in this 4-minute block?

A e^-4

B e^-16

C 16 × e^-16

D 0

💡 Short Explanation

For a Poisson distribution, [ P(X=0)=e^{-\lambda}. ] Here, λ = 16. Hence, the required probability is (e^{-16}).

✅ Correct Answer Explanation

The Poisson probability formula is [ P(X=k)=\frac{\lambda^k e^{-\lambda}}{k!} ] For zero customers, [ k=0 ] Thus, [ P(X=0)=\frac{16^0e^{-16}}{0!}=e^{-16} ] Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A) e^-4 → Uses the wrong λ.
Option C) 16 × e^-16 → Corresponds to (P(X=1)).
Option D) 0 → Probability is small but not zero.

🎯 Strategy

Used

Substitution

Application: Substitute (k=0) into the Poisson probability formula.

Final Logic: (P(0)=e^{-16}), so Option B is correct.

🧠 Memory Trick

"Zero Events → Just (e^{-λ})."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Poisson Probability Formula

12 Integrating over the traffic flow where λ = 3.2 bicycles/hour, what is the probability that 3 OR MORE riders use the track in an hour? (Given P(X≤2) = 0.381)

A 0.381

B 0.500

C 0.619

D 0.820

💡 Short Explanation

Use the complement rule. (P(X\ge3)=1-P(X\le2)). Substitute the given cumulative probability.

✅ Correct Answer Explanation

Given, [ P(X\le2)=0.381 ] Using the complement rule, [ P(X\ge3)=1-P(X\le2) ] Therefore, [ P(X\ge3)=1-0.381=0.619 ] Hence, Option C is correct. Option A: Probability of at most two riders. Option B: Incorrect computation. Option C: Correct complement. Option D: Incorrect subtraction.

❓ Why Other Options Are Incorrect

Option A) 0.381 → Represents (P(X\le2)), not (P(X\ge3)).
Option B) 0.500 → Not obtained from the given information.
Option D) 0.820 → Incorrect complement value.

🎯 Strategy

Used

Substitution

Application: Apply the complement rule directly using the given cumulative probability.

Final Logic: Since (P(X\ge3)=1-0.381=0.619), Option C is correct.

🧠 Memory Trick

"At Least = 1 − At Most."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Complement Rule in Poisson Distribution

📄 Passage

Let us evaluate a traffic engineer recording bicycle riders on a track to understand predicting rare events. He observes an average of 3.2 riders approaching every hour (λ = 3.2). The goal is to find the probability that 3 or more bicycle riders will use the track within an hour. Because there is no upper limit to 'k' in a Poisson distribution (it can technically go to infinity), this specific upper probability cannot be calculated directly through basic summation. Instead, we use the rule of complement. We first calculate the discrete probability of 2 or fewer riders:

[
P(X\le2)=P(X=0)+P(X=1)+P(X=2)
]
Once evaluated, we subtract this total from 1, yielding
[
P(X\ge3)=1-P(X\le2)
]
Based on the passage's logic regarding sums, why can't the probability P(X ≥ 3) be calculated directly by summing P(X=k)?

A Because the upper limit of k extends to infinity.

B Because the mean λ is a fraction.

C Because events in Poisson are dependent.

D Because the standard deviation is unknown.

💡 Short Explanation

A Poisson random variable has no finite upper limit. Directly summing probabilities from 3 to infinity is impractical. The complement rule provides a simpler calculation.

✅ Correct Answer Explanation

In a Poisson distribution, the random variable can take values [ 0,1,2,3,\ldots,\infty ] Therefore, [ P(X\ge3)=P(3)+P(4)+P(5)+\cdots ] Since this is an infinite sum, it is easier to use [ P(X\ge3)=1-P(X\le2) ] Thus, Option A is correct. Option A: Correct because the upper limit extends to infinity. Option B: The value of λ does not prevent direct summation. Option C: Poisson events are assumed independent. Option D: Standard deviation is not required for this reasoning.

❓ Why Other Options Are Incorrect

Option B) → λ being fractional has no effect on the complement rule.
Option C) → Poisson events are independent.
Option D) → Standard deviation is unrelated to calculating cumulative probabilities.

🎯 Strategy

Used

Conceptual Recall

Application: Recall that Poisson distributions have infinitely many possible values.

Final Logic: Since the upper limit is infinite, the complement rule is preferred, making Option A correct.

🧠 Memory Trick

"Poisson Tail → Use Complement."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Poisson Distribution and Complement Rule

📄 Passage

[
P(X\le2)=P(X=0)+P(X=1)+P(X=2)
]
Once evaluated, we subtract this total from 1, yielding
[
P(X\ge3)=1-P(X\le2)
]
What fundamental mathematical principle is applied in the passage to successfully solve for the probability of 3 or more riders?

A Multiplication theorem of probability

B Bayes' Theorem

C The Rule of Complement (Total probability is 1)

D Infinite Geometric Series sum

💡 Short Explanation

Total probability equals 1. The desired probability is obtained by subtracting the complementary probability. This is known as the complement rule.

✅ Correct Answer Explanation

The passage applies the identity [ P(X\ge3)=1-P(X\le2) ] This follows directly from the probability axiom that the probabilities of complementary events sum to 1. Therefore, Option C is correct. Option A: Not used. Option B: Unrelated. Option C: Correct complement rule. Option D: Not applicable.

❓ Why Other Options Are Incorrect

Option A) → Multiplication theorem is not involved.
Option B) → Bayes' theorem is unrelated.
Option D) → No geometric series is used.

🎯 Strategy

Used

Formula Recall

Application: Recall the complement rule for cumulative probabilities.

Final Logic: Since total probability equals 1, Option C is correct.

🧠 Memory Trick

"At Least = 1 − At Most."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Complement Rule

15 In the bicycle riders problem where λ = 3.2, what is the analytical expansion used to calculate P(X=2)?

A (3.2² × e^-3.2) / 2

B (3.2 × e^-3.2) / 2

C (3.2² × e^-3.2) / 4

D (e^-3.2) / 2

💡 Short Explanation

Use the Poisson probability formula. Substitute λ = 3.2 and k = 2. Since (2! = 2), the denominator is 2.

✅ Correct Answer Explanation

The Poisson formula is [ P(X=k)=\frac{\lambda^k e^{-\lambda}}{k!} ] For (k=2), [ P(X=2)=\frac{(3.2)^2e^{-3.2}}{2!} =\frac{(3.2)^2e^{-3.2}}{2} ] Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B) → Uses λ instead of λ².
Option C) → Incorrect denominator.
Option D) → Omits λ².

🎯 Strategy

Used

Substitution

Application: Substitute λ and k into the Poisson formula.

Final Logic: Using (k=2) gives Option A.

🧠 Memory Trick

"Poisson = λᵏe⁻ˡ / k!"

📚 Class XI – Mathematics – Chapter: Probability – Topic: Poisson Distribution

16 Given the river flood model (λ = 2), what is the calculated probability of 3 OR LESS overflow floods in 10 years? (Given P(X=0)=0.14, P(X=1)=0.27, P(X=2)=0.27, P(X=3)=0.18)

A 0.54

B 0.68

C 0.86

D 0.95

💡 Short Explanation

Add probabilities from 0 through 3. This gives (P(X\le3)). Sum = 0.86.

✅ Correct Answer Explanation

[ P(X\le3)=P(0)+P(1)+P(2)+P(3) ] [ =0.14+0.27+0.27+0.18 =0.86 ] Thus, Option C is correct.

❓ Why Other Options Are Incorrect

Option A) → Incomplete sum.
Option B) → Incorrect addition.
Option D) → Larger than the correct cumulative probability.

🎯 Strategy

Used

Substitution

Application: Add the given probabilities directly.

Final Logic: The cumulative probability equals 0.86, making Option C correct.

🧠 Memory Trick

"Less Than = Add Successively."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Cumulative Probability

17 For the airport arriving passengers problem, evaluating the last 50 minutes gives λ = 25. What is the probability expression of exactly 2 arrivals occurring in this 50-minute interval?

A (25² × e^-25) / 2

B (2²⁵ × e^-2) / 25!

C (25 × e^-25) / 2

D 25 × e^-2

💡 Short Explanation

Apply the Poisson formula with λ = 25 and k = 2. Since (2! = 2), simplify accordingly. The required expression matches Option A.

✅ Correct Answer Explanation

[ P(X=2)=\frac{25^2e^{-25}}{2!} =\frac{25^2e^{-25}}{2} ] Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B) → Incorrect formula.
Option C) → Uses λ instead of λ².
Option D) → Omits factorial and λ².

🎯 Strategy

Used

Formula Recall

Application: Recall the Poisson probability formula.

Final Logic: Substituting λ = 25 and k = 2 gives Option A.

🧠 Memory Trick

"Square λ when k = 2."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Poisson Distribution

18 In the same airport scenario, solving for the probability of 2 or more arrivals (P(k ≥ 2)) in the last 50 minutes (λ = 25) requires evaluating which complement expression?

A 1 − [e^-25]

B 1 − [(e^-25) + (25 × e^-25)]

C 1 − [(25² × e^-25) / 2]

D (e^-25) + (25 × e^-25)

💡 Short Explanation

Use the complement rule. Subtract (P(0)) and (P(1)) from 1. This gives (P(X\ge2)).

✅ Correct Answer Explanation

[ P(X\ge2)=1-P(X\le1) ] where [ P(0)=e^{-25} ] and [ P(1)=25e^{-25} ] Thus, [ P(X\ge2)=1-[e^{-25}+25e^{-25}] ] Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A) → Omits (P(1)).
Option C) → Uses (P(2)) instead of the complement.
Option D) → Gives only the complement event.

🎯 Strategy

Used

Substitution

Application: Apply the complement rule using (P(0)) and (P(1)).

Final Logic: (P(X\ge2)=1-P(X\le1)), making Option B correct.

🧠 Memory Trick

"Greater or Equal → 1 − Smaller Cases."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Complement Rule

19 Using the mortality rate Poisson model where λ = 2.8, what is the exact mathematical expression for evaluating the probability of exactly 2 deaths?

A (2.8 × e^-2.8)

B (2.8² × e^-2.8) / 2!

C (2.8² × e^-2) / 2.8!

D (2^2.8 × e^-2) / 2!

💡 Short Explanation

Apply the Poisson formula. Use λ = 2.8 and k = 2. Keep the factorial as 2!.

✅ Correct Answer Explanation

[ P(X=2)=\frac{(2.8)^2e^{-2.8}}{2!} ] Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A) → Corresponds to (P(X=1)).
Option C) → Incorrect exponential and factorial.
Option D) → Incorrect formula.

🎯 Strategy

Used

Formula Recall

Application: Substitute λ and k into the Poisson formula.

Final Logic: The correct expression is Option B.

🧠 Memory Trick

"Poisson = λᵏ/k! × e⁻ˡ."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Poisson Distribution

20 A customer care company receives an average of 4.5 calls every 5 minutes. If evaluating the probability of getting exactly 1 call in this 5-minute interval, the formula simplifies to:

A 4.5 × e^-4.5

B e^-4.5

C (4.5)² / 2

D 1 − e^-4.5

💡 Short Explanation

Use the Poisson formula with (k=1). Since (1!=1), the expression simplifies directly. The result is (4.5e^{-4.5}).

✅ Correct Answer Explanation

For (k=1), [ P(X=1)=\frac{\lambda e^{-\lambda}}{1!} ] Substituting (\lambda=4.5), [ P(X=1)=4.5e^{-4.5} ] Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B) → Represents (P(X=0)).
Option C) → Omits the exponential factor.
Option D) → Represents (P(X\ge1)).

🎯 Strategy

Used

Substitution

Application: Substitute (k=1) into the Poisson probability formula.

Final Logic: Since (P(1)=\lambda e^{-\lambda}), Option A is correct.

🧠 Memory Trick

"One Event → λe⁻ˡ."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Poisson Distribution Formula

CUET UG Applied Mathematics Booster Test 3 - Poisson Distribution

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