The length of time for one person to be served at a restaurant is exponentially distributed with an expected time of
5
5 minutes. If
5
5 persons arrived at the restaurant then what is the probability that at least
2
2 of them will be served in more than
8
8 minutes? (Assume that each person is served independently

To solve this problem, we first need to understand the properties of the exponential distribution and the situation described.

### Step 1: Define the Exponential Distribution
The time \( T \) for one person to be served is exponentially distributed with an expected value \( E(T) = 5 \) minutes. For an exponential distribution, the probability density function (PDF) is given by:

\[
f(t) = \lambda e^{-\lambda t} \quad \text{for } t \geq 0,
\]

where \( \lambda = \frac{1}{\text{Expected Time}} = \frac{1}{5} \) (since the expected time is 5 minutes).

### Step 2: Calculate the Probability That a Person is Served in More Than 8 Minutes
The probability that one person is served in more than 8 minutes is:

\[
P(T > 8) = 1 - P(T \leq 8)
\]

For the exponential distribution, \( P(T \leq t) \) is given by the cumulative distribution function (CDF):

\[
P(T \leq t) = 1 - e^{-\lambda t}
\]

Thus:

\[
P(T > 8) = e^{-\lambda \cdot 8} = e^{-\frac{8}{5}} = e^{-1.6}
\]

### Step 3: Probability That at Least 2 Out of 5 Persons Are Served in More Than 8 Minutes
Let \( X \) be the number of persons served in more than 8 minutes. \( X \) follows a binomial distribution since each person is served independently:

\[
X \sim \text{Binomial}(n = 5, p = e^{-1.6})
\]

We need to find \( P(X \geq 2) \), which is the probability that at least 2 persons are served in more than 8 minutes. This can be calculated as:

\[
P(X \geq 2) = 1 - P(X < 2)
\]

\( P(X < 2) \) is the probability that fewer than 2 people are served in more than 8 minutes, which includes:

\[
P(X < 2) = P(X = 0) + P(X = 1)
\]

Where:

\[
P(X = 0) = \binom{5}{0} (e^{-1.6})^0 (1 - e^{-1.6})^5 = (1 - e^{-1.6})^5
\]

\[
P(X = 1) = \binom{5}{1} (e^{-1.6})^1 (1 - e^{-1.6})^4 = 5e^{-1.6}(1 - e^{-1.6})^4
\]

Thus:

\[
P(X \geq 2) = 1 - \left[(1 - e^{-1.6})^5 + 5e^{-1.6}(1 - e^{-1.6})^4\right]
\]

### Step 4: Final Calculation
Let's calculate the numerical value:

1. \( e^{-1.6} \approx 0.2019 \)
2. \( 1 - e^{-1.6} \approx 0.7981 \)

Now, compute the probabilities:

\[
P(X = 0) \approx (0.7981)^5 \approx 0.3282
\]

\[
P(X = 1) \approx 5 \times 0.2019 \times (0.7981)^4 \approx 0.4136
\]

\[
P(X < 2) \approx 0.3282 + 0.4136 = 0.7418
\]

Finally:

\[
P(X \geq 2) = 1 - 0.7418 = 0.2582
\]

So, the probability that at least 2 out of 5 persons will be served in more than 8 minutes is approximately **0.2582** or **25.82%**.

---

Would you like more details or have any questions?

Here are five related questions:
1. How would the probability change if the expected service time was 10 minutes instead?
2. What is the probability that exactly 3 out of 5 persons are served in more than 8 minutes?
3. Can you find the expected number of people served in more than 8 minutes?
4. How does the binomial distribution apply to this type of problem?
5. What is the significance of the memoryless property in the exponential distribution?

**Tip:** The exponential distribution is memoryless, meaning the probability of an event occurring in the future is independent of how much time has already elapsed.

Calculate the probability that at least 2 out of 5 restaurant customers will be served in more than 8 minutes. Understand the application of exponential and binomial distributions in probability theory.

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