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.