Calculating the Probability of At Most ‘X‘ Events Occurring

1. Use the formula to calculate each probability up to and including x: [latex]P(X\le x) = P(X=0)+P(X=1)+...+P(X=x)[/latex]
2. Use cumulative = [latex]TRUE[/latex] in Excel: [latex]P(X\le x)[/latex] = POISSON.DIST([latex]x[/latex], [latex]\lambda[/latex], TRUE)

Which is better? It is often much quicker to use Excel’s POISSON.DIST than the formula.

Calculating the Probability of More Than ‘X‘ Events Occurring

1. Use the formula and a complement to calculate: [latex]P(X\gt x) = 1-[P(0)+P(1)+...+P(x)][/latex]
2. Use a complement and [latex]TRUE[/latex] in Excel: [latex]P(X\gt x) =1−[/latex]POISSON.DIST([latex]x[/latex], [latex]\lambda[/latex], TRUE)

Why use a complement? Since there is no upper limit to the number of events that can occur, [latex]P(X\gt x) = P(x+1)+P(x+2)+...+P(\infty)[/latex]. We cannot perform this calculation without using limit theory. Instead, we use a complement of up to the probability of [latex]x[/latex] events occurring.

Calculating the Probability of At Least ‘X‘ Events Occurring

1. Use the formula and a complement: [latex]P(X\ge x) = 1-[P(0)+P(1)+...+P(x-1)][/latex]
2. Use a complement and [latex]TRUE[/latex] in Excel: [latex]P(X\ge x)[/latex] = 1− POISSON.DIST([latex]x-1[/latex], [latex]\lambda[/latex], TRUE)

Why do we use [latex]x-1[/latex] instead of [latex]x[/latex]? Since [latex]x[/latex] is included in the interval, it needs to be excluded when taking the complement. Ie: [latex]P(X\ge x) = 1-P(X \lt x) = 1-P(X \le x-1)[/latex]

‘At MOST’ Example (VIDEO)

• It is important to stress that we must always match the time units.
• In the example from the previous section, the average arrival rate was given per minute.
• The question’s time interval was also one minute: “What is the probability of 5 customers arriving in any given minute.”
• So, no adjustment of lambda ([latex]\lambda[/latex]) was needed since the time intervals matched.
• See the next example to see where an adjustment of [latex]\lambda[/latex] is needed.

Example 32.1

Problem Setup: The Canadian Anti-Fraud Centre received an average of 3,578 reports of fraud per month so far in 2024, with losses totaling $123,000,000 in the first quarter of 2024 alone!

Question: If we assume that the number of frauds reported to the Canadian Anti-Fraud Centre follows a Poisson distribution, what is the probability of there being at most 3,578 frauds reported in the month of May?

Solutions: Click here to download the Excel solutions shown in the video below:

Conclusion: We use =POISSON.DIST(3578, 3578, TRUE). This gives a 50.44% chance that the number of reports received by the Anti-Fraud Centre in the month of May do not exceed the monthly average for 2024 to date.

‘More than’ Example (EXERCISE)

Let us now review how to calculate the probability of more than [latex]x[/latex] events occurring during a certain time frame in the exercise in the next example.

Example 32.2

Problem Setup: The Canadian Anti-Fraud Centre reports an average of 2,636 victims of fraud per month so far in 2024. They reported a total of 41,988 victims in 2023 (or 3,499 victims per month).  Let us assume that the number of victims of fraud in 2024 follows a Poisson distribution with an average of 2,636 victims per month.

Question: What is the probability of there being more than 3,499 victims of fraud in any given month in 2024? Ie: What is the probability that any given month in 2024, the number of victims of fraud will exceed the monthly average for 2023?

You Try: Solve the above problem by placing the values in the correct positions below:

‘At Least’ Example (Video)

• Finally, we will practice calculating the probability of ‘at least’ [latex]x[/latex] events in a certain time-frame.
• See the next example for a video walk-through of how to calculate [latex]P(X \ge x)[/latex] in Excel.

Example 32.3

Problem Setup: There are, on average, 28,000 bankruptcies filed in Canada per year according to the Office of the Superintendent of Bankruptcy in Canada.

Question: If the number of bankruptcies follows a Poisson distribution, what is the probability of at least 2,400 bankruptcies getting filed in any given month in Canada?

Solution: Use = 1− POISSON.DIST(2399, 28000/12, TRUE). Click here to download the Excel solutions shown below:

Key Takeaways (EXERCISE)

Your Own Notes (EXERCISE)

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