19.X.6. Solved Examples

Example 1: If the average number of emails received in an hour is 5, what is the probability that exactly 3 emails are received in a particular hour?

Solution: Here, 𝜆 = 5, 𝑘 = 3

𝑃(𝑋 = 3) = \frac{e^{-5} imes 5^3}{3!} = \frac{e^{-5} imes 125}{6} \approx 0.1404

In this example, we are using the Poisson distribution to find the probability of receiving exactly 3 emails in an hour when the average rate (λ) is 5 emails per hour. To solve it, we use the formula for the Poisson probability mass function:

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

Substituting the values:

• λ (the average rate of emails) = 5
• k (the number of emails we want the probability for) = 3

So, we compute:

• Calculate 𝑒^{-5} (using the approximate value for e)
• Raise λ to the power of k (5^3 = 125)
• Compute k! (3! = 6)

Finally, we plug these into our formula to find the probability, which comes out to approximately 0.1404, indicating there is about a 14.04% chance of receiving exactly 3 emails in that hour.

Imagine you're at a coffee shop where customers arrive randomly throughout the hour. If, on average, 5 customers arrive per hour, you can use this statistical approach to determine the likelihood of having exactly 3 customers in the next hour. Just like counting the emails, it's useful for making predictions based on the average rate of events.

Example 2: A manufacturing unit produces a defect on average every 2 meters. Find the probability that there will be no defect in a 4-meter length.

Solution: Rate = 1 defect / 2 meters ⇒ 𝜆 = 4/2 = 2

𝑃(𝑋 = 0) = \frac{e^{-2} \cdot 2^0}{0!} = e^{-2} \approx 0.1353

In this example, we want to calculate the probability of finding no defects in a 4-meter length of product when defects occur at an average rate of 1 defect every 2 meters. First, we determine λ (the average number of defects in the interval of interest).

1. Calculate λ for the 4-meter length: since defects occur every 2 meters, within the 4-meter length, we expect 2 defects (λ = 2).
2. We want to find the probability of 0 defects (k = 0). Using the same Poisson formula:

$P(X = 0) = \frac{e^{-\lambda} \lambda^k}{k!}$

3. Substitute the values:

• λ = 2
• k = 0

Here, we have:

• e^{-2} (calculating the exponential part)
• 2^0 = 1
• 0! = 1

Plugging these numbers in yields the probability of finding no defects; it comes out to about 0.1353, or a 13.53% chance of having no defects in that section of product.

Think about measuring fruit in a shipment. If, on average, you find one bruised apple for every two boxes, and you check four boxes, you can calculate the likelihood of finding no bruised apples at all. Using the same Poisson distribution, you can estimate your chances, which helps in quality inspection scenarios.

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