Question

The length of time for one individual to be served at a cafeteria is an exponential random variable with mean of 6 minutes. Assume a person has waited for at least 4 minutes to be served. What is the probability that the person will need to wait at least 9 minutes total

Answer 1

Answer:

43.46% probability that the person will need to wait at least 9 minutes total

Step-by-step explanation:

To solve this question, we need to understand conditional probability and the exponential distribution.

Conditional probability:

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

Expontial distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

In this question:

Event A: Waited at least 4 minutes.

Event B: Waiting at least 9 minutes.

The length of time for one individual to be served at a cafeteria is an exponential random variable with mean of 6 minutes.

This means that $m = 6, \mu = \frac{1}{6}$

Probability of waiting at least 4 minutes.

$P(A) = P(X \geq 4) = P(X > 4)$

$P(A) = P(X > 4) = e^{-\frac{4}{6}} = 0.5134$

Intersection:

The intersection between a waiting time of at least 4 minutes and a waiting time of at list 9 minutes is a waiting time of 9 minutes. So

$P(A \cap B) = P(X > 9) = e^{-\frac{9}{6}} = 0.2231$

What is the probability that the person will need to wait at least 9 minutes total

$P(B|A) = \frac{0.2231}{0.5134} = 0.4346$

43.46% probability that the person will need to wait at least 9 minutes total