Their manager decides to pool Veda and Jomo’s calls. Now, they receive calls at a rate of 1.5+1.2 = 2.7 in 5 minutes. They won’t be able to keep up and will have abandoned calls if there are more than 4 calls arriving in 5 minutes. Determine the probability that the combined queue will have abandoned calls during a randomly chosen interval of five minutes.


This problem involves a Poisson process since calls are arriving randomly over time, and we need to determine the probability of having more than 4 calls arriving in 5 minutes. The arrival rate of calls is the sum of Veda's and Jomo's individual rates, i.e., \( 1.5 + 1.2 = 2.7 \) calls per 5 minutes.

### Step 1: Understand the Poisson distribution
The number of calls in a Poisson process follows the Poisson distribution. The probability mass function for the Poisson distribution is given by:

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

where:
- \( \lambda \) is the average number of events (in this case, the average number of calls in 5 minutes),
- \( k \) is the number of events (calls),
- \( e \) is the base of the natural logarithm (approximately 2.71828).

In this case, \( \lambda = 2.7 \) calls per 5 minutes.

### Step 2: Probability of more than 4 calls
We need to find the probability that more than 4 calls arrive in 5 minutes. This is equivalent to calculating:

\[
P(X > 4) = 1 - P(X \leq 4)
\]

To calculate \( P(X \leq 4) \), we need to sum the probabilities for \( X = 0, 1, 2, 3, 4 \).

\[
P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
\]

Using the Poisson formula for each \( k \):

\[
P(X = k) = \frac{2.7^k e^{-2.7}}{k!}
\]

We will now calculate each term and sum them.

### Step 3: Calculate the probabilities
We will calculate the probabilities step by step for \( k = 0, 1, 2, 3, 4 \) and sum them up to get \( P(X \leq 4) \), then subtract from 1 to get \( P(X > 4) \).The probability that more than 4 calls will arrive during a randomly chosen 5-minute interval, leading to abandoned calls, is approximately **0.137** or **13.7%**.

Would you like more details on how the calculations were done or have any further questions?

Here are 5 related questions to expand on this topic:
1. How does the Poisson distribution change if the call rate increases to 3.5 per 5 minutes?
2. What is the probability of receiving exactly 5 calls in 5 minutes?
3. How would the probability of more than 4 calls change if the time interval was 10 minutes instead of 5?
4. What is the expected number of calls in a 10-minute interval if the rate remains 2.7 per 5 minutes?
5. How would a different queueing strategy impact the probability of abandoned calls?

**Tip:** In Poisson processes, the time interval and the rate are crucial factors. If you double the time, the expected number of events also doubles.

Their manager decides to pool Veda and Jomo’s calls. Now, they receive calls at a rate of 1.5+1.2 = 2.7 in 5 minutes. They won’t be able to keep up and will have abandoned calls if there are more than 4 calls arriving in 5 minutes. Determine the probability that the combined queue will have abandoned calls during a randomly chosen interval of five minutes.

Learn how to calculate the probability of abandoned calls in a Poisson process where the combined call rate of Veda and Jomo is 2.7 calls per 5 minutes. This detailed solution includes the use of the Poisson distribution to determine the probability of more than 4 calls arriving in 5 minutes, leading to abandoned calls. Suitable for university-level statistics students.

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