Question

asked Nov 8, 2021 147k views

1 Answer

Veky · Answer 1 · 2021-11-14T17:04:30+0000

Answer:

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

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

$P(B|A) = (P(A \cap B))/(P(A))$

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
$\mu = (1)/(m)$ 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
$m = 6, \mu = (1)/(6)$

Probability of waiting at least 4 minutes.

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

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

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

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

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

P(B|A) = (0.2231)/(0.5134) = 0.4346

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

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