Computing Event Likelihood Using the Poisson Model

EASY10 pts

In probability theory and statistics, the Poisson distribution is a fundamental discrete probability distribution that models the likelihood of a given number of events occurring within a fixed interval of time, space, or any other continuous domain. This distribution is particularly powerful when events occur independently at a constant average rate.

The Poisson distribution is characterized by a single parameter λ (lambda), which represents the expected number of occurrences (mean rate) in the specified interval. The probability of observing exactly k events is given by the Poisson Probability Mass Function (PMF):

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

Where:

k is the number of events (a non-negative integer: 0, 1, 2, 3, ...)
λ (lambda) is the average rate of occurrence (a positive real number)
e is Euler's number (approximately 2.71828...)
k! is the factorial of k

Key Properties of the Poisson Distribution:

The mean and variance are both equal to λ
As λ increases, the distribution approaches a normal distribution
The distribution is right-skewed for small λ values and becomes more symmetric as λ grows
It is often used when events are rare relative to the opportunities for them to occur

Your Task: Implement a Python function that computes the probability of observing exactly k events given the average rate λ using the Poisson distribution formula. Your function should return the computed probability rounded to 5 decimal places for precision.

Example

Input

k = 3, lam = 5

Output

0.14037

Explanation

With λ = 5 (expected 5 events on average) and k = 3 (observing exactly 3 events):

• λ^k = 5^3 = 125 • e^(-λ) = e^(-5) ≈ 0.006738 • k! = 3! = 6 • P(X = 3) = (125 × 0.006738) / 6 ≈ 0.14037

There is approximately a 14.04% chance of observing exactly 3 events when the average rate is 5.

Example

Input

k = 0, lam = 2

Output

0.13534

Explanation

With λ = 2 (expected 2 events on average) and k = 0 (observing zero events):

• λ^k = 2^0 = 1 (any number raised to power 0 is 1) • e^(-λ) = e^(-2) ≈ 0.13534 • k! = 0! = 1 (by definition, 0 factorial equals 1) • P(X = 0) = (1 × 0.13534) / 1 ≈ 0.13534

There is approximately a 13.53% probability of observing no events when the expected average is 2. This represents the probability that in a given interval, nothing occurs despite expecting 2 occurrences on average.

Example

Input

k = 5, lam = 5

Output

0.17547

Explanation

With λ = 5 (expected 5 events on average) and k = 5 (observing exactly 5 events):

• λ^k = 5^5 = 3125 • e^(-λ) = e^(-5) ≈ 0.006738 • k! = 5! = 120 • P(X = 5) = (3125 × 0.006738) / 120 ≈ 0.17547

When k equals λ, we are calculating the probability of observing exactly the expected number of events. This is the mode (most likely value) for integer λ, with approximately 17.55% probability.

Accepted0/0·0% Acceptance

Constraints

0 ≤ k ≤ 100 (number of events must be a non-negative integer)
0 < lam ≤ 100 (lambda must be positive)
The result should be rounded to exactly 5 decimal places
Use appropriate mathematical libraries for factorial and exponential calculations
Handle the edge case where k = 0 correctly (0! = 1)

Code

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Solutions

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Test Cases3

Results

Submissions

k =

lam =

Computing Event Likelihood Using the Poisson Model

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Computing Event Likelihood Using the Poisson Model

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