Math Problem Statement

Which is the correct option

The length of time for one person to be served at a restaurant is exponentially distributed with an expected time of 4 minutes. If 5 persons arrived at the restaurant, what is the probability that at least 2 of them will be served in more than 6 minutes? Assume that each person is served independently.

Solution

To solve this problem systematically, let's analyze it step by step.


Given Information

  1. Service Time Distribution: The service time TT follows an exponential distribution with an expected value E(T)=4E(T) = 4 minutes.
  2. Rate Parameter λ\lambda: The exponential distribution has rate λ\lambda, which is related to the mean as: λ=1mean=14.\lambda = \frac{1}{\text{mean}} = \frac{1}{4}.
  3. Time Threshold t=6t = 6 minutes: We are tasked to find the probability that at least 2 people out of 5 are served in more than 6 minutes.
  4. Independence: The service times of the 5 individuals are independent.

Step 1: Probability for One Person

For an exponential distribution: P(T>t)=eλt.P(T > t) = e^{-\lambda t}. Substitute λ=14\lambda = \frac{1}{4} and t=6t = 6: P(T>6)=e64=e1.5.P(T > 6) = e^{-\frac{6}{4}} = e^{-1.5}. Thus, the probability that one person is served in more than 6 minutes is e1.5e^{-1.5}.

Let p=e1.5p = e^{-1.5} and q=1pq = 1 - p (probability that service is completed within 6 minutes).


Step 2: Binomial Distribution

The number of people served in more than 6 minutes out of 5 follows a binomial distribution: XBinomial(n=5,p=e1.5).X \sim \text{Binomial}(n = 5, p = e^{-1.5}). We are tasked with finding the probability that at least 2 of them are served in more than 6 minutes. Mathematically: P(X2)=1P(X=0)P(X=1).P(X \geq 2) = 1 - P(X = 0) - P(X = 1).


Step 3: Compute P(X=0)P(X = 0) and P(X=1)P(X = 1)

For a binomial distribution: P(X=k)=(nk)pkqnk.P(X = k) = \binom{n}{k} p^k q^{n-k}.

  • For X=0X = 0: P(X=0)=(50)p0q5=q5.P(X = 0) = \binom{5}{0} p^0 q^5 = q^5.
  • For X=1X = 1: P(X=1)=(51)p1q4=5pq4.P(X = 1) = \binom{5}{1} p^1 q^4 = 5 p q^4.

Step 4: Combine Results

Now, compute P(X2)P(X \geq 2): P(X2)=1P(X=0)P(X=1).P(X \geq 2) = 1 - P(X = 0) - P(X = 1). Substitute the expressions: P(X2)=1q55pq4.P(X \geq 2) = 1 - q^5 - 5p q^4. Recall that p=e1.5p = e^{-1.5} and q=1e1.5q = 1 - e^{-1.5}.


Step 5: Identify the Correct Option

From the options:

  1. First option matches: 1(1e1.5)5[5e1.5(1e1.5)4].1 - \left( 1 - e^{-1.5} \right)^5 - \left[ 5 e^{-1.5} \left( 1 - e^{-1.5} \right)^4 \right]. This matches our derived formula for P(X2)P(X \geq 2).

Final Answer:

The correct option is the first one.


Do you want a more detailed explanation of any step or related concepts?


Related Questions:

  1. What is the cumulative distribution function (CDF) for the exponential distribution?
  2. How do we calculate the mean and variance for an exponential distribution?
  3. What is the significance of the binomial distribution in probability theory?
  4. How do you generalize the probability of "at least kk" successes in a binomial distribution?
  5. Can you derive the moment generating function for an exponential distribution?

Tip: For exponential distributions, the memoryless property states that the probability of waiting an additional time does not depend on how much time has already passed.

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Math Problem Analysis

Mathematical Concepts

Exponential Distribution
Binomial Distribution
Probability Theory

Formulas

P(T > t) = e^{-λt}
P(X = k) = C(n, k) p^k (1-p)^{n-k}
P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)

Theorems

Memoryless Property of Exponential Distribution
Binomial Distribution Probability Formula

Suitable Grade Level

Undergraduate or Advanced High School Level