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Animated Solution for Mathematics - Probability: At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of 5 phone calls during 10 minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is

Q: Why is $0!$ equal to $1$?

It feels counterintuitive, doesn't it? In mathematics, $0!$ is defined as $1$ to maintain the consistency of the factorial function, specifically for the gamma function and the Poisson formula itself. Without this definition, the formula would break down!

Select Answer:

Visualized Solution

Identifying the Poisson Parameter $λ$

Average number of calls in $10$ minutes: $λ = 5$
Random variable $X$ : Number of phone calls in a $10$ -minute interval
Since calls occur independently at a constant average rate, $X$ follows a Poisson distribution.

The Poisson Probability Mass Function

Poisson Probability Mass Function: $P (X = r) = \frac{e ^{- λ} λ ^{r}}{r !}$
Where $e \approx 2.718$ is Euler's constant.
$λ$ is the average number of occurrences.
$r$ is the actual number of occurrences we want to find the probability for.

Visualizing the Distribution

The probability varies for different values of $X$ ( $0, 1, 2, \dots$ ).
The sum of all probabilities for $X = 0$ to $\infty$ is exactly $1$ .

Defining 'At Most One' Call

Event: 'At most one phone call' means $X \leq 1$ .
This includes exactly two mutually exclusive cases: $X = 0$ or $X = 1$ .
Mathematical expression: $P (X \leq 1) = P (X = 0) + P (X = 1)$ .

Setting up $P (X = 0)$

Substitute $λ = 5$ and $r = 0$ into the Poisson formula.
$P (X = 0) = \frac{e ^{- 5} \cdot 5 ^{0}}{0 !}$

Calculating $P (X = 0)$

Recall the algebraic rules: $5^{0} = 1$ and $0! = 1$ .
$P (X = 0) = \frac{e ^{- 5} \cdot 1}{1}$
$P (X = 0) = e^{- 5}$

Setting up $P (X = 1)$

Substitute $λ = 5$ and $r = 1$ into the Poisson formula.
$P (X = 1) = \frac{e ^{- 5} \cdot 5 ^{1}}{1 !}$

Calculating $P (X = 1)$

Recall the algebraic rules: $5^{1} = 5$ and $1! = 1$ .
$P (X = 1) = \frac{e ^{- 5} \cdot 5}{1}$
$P (X = 1) = 5 e^{- 5}$

Summing the Probabilities

Combine the results for $X = 0$ and $X = 1$ .
$P (X \leq 1) = e^{- 5} + 5 e^{- 5}$

Final Simplification

Factor out the common term $e^{- 5}$ .
$P (X \leq 1) = e^{- 5} (1 + 5) = 6 e^{- 5}$
Rewrite using positive exponents: $P (X \leq 1) = \frac{6}{e ^{5}}$
Final Answer: The probability is $\frac{6}{e ^{5}}$ .

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The Sigma Insight: Random Variables and Probability Distributions

Solution Diagram

Analyzing the Setup

In this scenario, we are modeling the arrival of telephone calls, which occur independently at a constant average rate. This process is governed by the Poisson distribution.

The parameter

λ

represents the average number of occurrences in the specified time interval. Given the problem constraints, we have:

λ = 5

The Master Equation

The probability mass function for the Poisson distribution, which gives the probability of exactly

r

occurrences, is defined as:

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

We are tasked with finding the probability of receiving at most one call in the ten-minute window. Mathematically, this is expressed as

P (X \leq 1)

Calculating the Probabilities

Since the Poisson distribution is discrete, the probability

P (X \leq 1)

is the sum of the probabilities of two mutually exclusive events:

X = 0

and

X = 1

P (X \leq 1) = P (X = 0) + P (X = 1)

First, we calculate the probability for

r = 0

P (X = 0) = \frac{e ^{- 5} \cdot 5 ^{0}}{0 !} = \frac{e ^{- 5} \cdot 1}{1} = e^{- 5}

Next, we calculate the probability for

r = 1

P (X = 1) = \frac{e ^{- 5} \cdot 5 ^{1}}{1 !} = \frac{e ^{- 5} \cdot 5}{1} = 5 e^{- 5}

Final Calculation

By summing these two results, we arrive at the total probability:

P (X \leq 1) = e^{- 5} + 5 e^{- 5} = 6 e^{- 5}

Expressed in fractional form, the final probability is:

\frac{6}{e ^{5}}

Common Traps & FAQs

Why do we use the Poisson distribution here instead of Binomial?

Great question! We use Poisson when we have a large number of trials (or a continuous time interval) where the probability of an individual event is very small, but the average rate is constant. Binomial is for a fixed number of trials

n

with a fixed probability

p

. Here, we have a rate, not a fixed number of trials.

What does 'at most one' actually mean?

It is a common point of confusion. 'At most one' means the number of calls

X

can be

0

1

. It is inclusive of the upper bound. If it said 'less than one', it would only be

0

. Always check if the boundary is included!

Why is

0!

equal to

1

It feels counterintuitive, doesn't it? In mathematics,

0!

is defined as

1

to maintain the consistency of the factorial function, specifically for the gamma function and the Poisson formula itself. Without this definition, the formula would break down!

Select Answer:

Visualized Solution

Identifying the Poisson Parameter $λ$

The Poisson Probability Mass Function

Visualizing the Distribution

Defining 'At Most One' Call

Setting up $P (X = 0)$

Calculating $P (X = 0)$

Setting up $P (X = 1)$

Calculating $P (X = 1)$

Summing the Probabilities

Final Simplification

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Master Equation

Calculating the Probabilities

Final Calculation

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Select Answer:

Visualized Solution

Identifying the Poisson Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

The Poisson Probability Mass Function

Visualizing the Distribution

Defining 'At Most One' Call

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=0)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=0)

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=1)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=1)

Summing the Probabilities

Final Simplification

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Master Equation

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Final Calculation

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