Poisson Probability Distribution: Examples and Practice Problems

The Poisson probability distribution models the number of times an event occurs randomly and independently within a fixed interval of time or space. It is a discrete distribution, meaning the random variable \( X \) can only take integer values.

Poisson Process: Definition and Conditions

A Poisson process must satisfy the following conditions:

  1. Events are discrete, random, and independent of each other.
  2. The average rate of occurrence \( \lambda \) is constant over the observed period.
  3. The probability of an event occurring in a given interval is the same for all intervals of equal length.
  4. Two events cannot occur at exactly the same time (mutually exclusive).

Examples of Poisson Processes

Poisson Probability Formula

If an event occurs an average of \( \lambda \) times over a period \( T \), the probability of it occurring exactly \( x \) times in that period is:

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

where:

Poisson distributions for different lambda values

Poisson vs. Binomial Distribution

Distribution Parameters Variable Range
Binomial Number of trials \( n \), probability of success \( p \) \( x = 0, 1, 2, \ldots, n \)
Poisson Average rate \( \lambda \) \( x = 0, 1, 2, \ldots, \infty \)

Poisson Distribution Examples with Solutions

Example 1: Computer Crashes

A computer crashes on average once every 4 months. Find the probability that over 4 months it will:

  1. Not crash at all.
  2. Crash exactly once.
  3. Crash exactly twice.
  4. Crash exactly three times.

Solution

Here, \( \lambda = 1 \) per 4 months.

  1. \[ P(X = 0) = \frac{e^{-1} \cdot 1^0}{0!} = e^{-1} \approx 0.36787 \]
  2. \[ P(X = 1) = \frac{e^{-1} \cdot 1^1}{1!} = e^{-1} \approx 0.36787 \]
  3. \[ P(X = 2) = \frac{e^{-1} \cdot 1^2}{2!} = \frac{e^{-1}}{2} \approx 0.18393 \]
  4. \[ P(X = 3) = \frac{e^{-1} \cdot 1^3}{3!} = \frac{e^{-1}}{6} \approx 0.06131 \]

Example 2: Call Center

A customer help center receives an average of 3.5 calls per hour. Find the probability that in a given hour it will:

  1. Receive at most 4 calls.
  2. Receive at least 5 calls.

Solution

Here, \( \lambda = 3.5 \) per hour.

  1. At most 4 calls means \( X = 0, 1, 2, 3, \text{or } 4 \).

    \[ \begin{aligned} P(X \le 4) &= P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) \\ &= \frac{e^{-3.5} \cdot 3.5^0}{0!} + \frac{e^{-3.5} \cdot 3.5^1}{1!} + \frac{e^{-3.5} \cdot 3.5^2}{2!} + \frac{e^{-3.5} \cdot 3.5^3}{3!} + \frac{e^{-3.5} \cdot 3.5^4}{4!} \\ &\approx 0.03020 + 0.10569 + 0.18496 + 0.21579 + 0.18881 = 0.72545 \end{aligned} \]

  2. At least 5 calls: \( P(X \ge 5) = 1 - P(X \le 4) = 1 - 0.72545 = 0.27455 \).

Example 3: Email Reception

A person receives on average 3 emails per hour. Find over a 2-hour period the probability that they will:

  1. Receive exactly 5 emails.
  2. Receive more than 2 emails.

Solution

The average over 2 hours is \( \lambda = 3 \times 2 = 6 \).

  1. \[ P(X = 5) = \frac{e^{-6} \cdot 6^5}{5!} \approx 0.16062 \]

  2. More than 2 emails: \( P(X > 2) = 1 - P(X \le 2) \).

    \[ \begin{aligned} P(X \le 2) &= P(X=0) + P(X=1) + P(X=2) \\ &= \frac{e^{-6} \cdot 6^0}{0!} + \frac{e^{-6} \cdot 6^1}{1!} + \frac{e^{-6} \cdot 6^2}{2!} \\ &\approx 0.00248 + 0.01487 + 0.04462 = 0.06197 \\ P(X > 2) &= 1 - 0.06197 = 0.93803 \end{aligned} \]

Example 4: Football Goals (Approximation)

The goals scored by a football player in 35 matches are:

Goals (\( x \))0123>3
Frequency (\( f \))1215620

Assuming goals follow a Poisson distribution, find the probability that in a given match the player:

  1. Scores exactly 1 goal.
  2. Scores at least 1 goal.

Solution

First, compute the average \( \lambda \):

\[ \lambda = \frac{\sum f \cdot x}{\sum f} = \frac{12(0) + 15(1) + 6(2) + 2(3)}{12 + 15 + 6 + 2} = \frac{33}{35} \approx 0.9429 \]
  1. \[ P(X = 1) = \frac{e^{-0.9429} \cdot 0.9429^1}{1!} \approx 0.36719 \]
  2. \[ P(X \ge 1) = 1 - P(X = 0) = 1 - \frac{e^{-0.9429} \cdot 0.9429^0}{0!} \approx 1 - 0.3897 = 0.6103 \]

Example 5: Defective Items Returned

Defective items returned daily over 100 days:

Defective Items (\( x \))01234>4
Frequency (\( f \))5020151050

Assuming a Poisson distribution, find the probability that on a given day:

  1. No defective item is returned.
  2. Three or more defective items are returned.

Solution

Compute \( \lambda \):

\[ \lambda = \frac{50(0) + 20(1) + 15(2) + 10(3) + 5(4)}{50 + 20 + 15 + 10 + 5} = \frac{100}{100} = 1 \]
  1. \[ P(X = 0) = \frac{e^{-1} \cdot 1^0}{0!} = e^{-1} \approx 0.36787 \]

  2. Three or more: \( P(X \ge 3) = 1 - P(X \le 2) \).

    \[ \begin{aligned} P(X \le 2) &= P(X=0) + P(X=1) + P(X=2) \\ &= e^{-1} + e^{-1} + \frac{e^{-1}}{2} = e^{-1} \left(1 + 1 + \frac{1}{2}\right) \\ &= 2.5 e^{-1} \approx 2.5 \times 0.36787 = 0.91968 \\ P(X \ge 3) &= 1 - 0.91968 = 0.08032 \end{aligned} \]

Additional Resources