Question

Six customers enter a three-floor restaurant. Each customer decides on which floor to have dinner. Assume that the decisions of different customers are independent, and that for each customer, each floor is equally likely. Find the probability that exactly one customer dines on the first floor.

Answer 1

Answer:

The probability that exactly one customer dines on the first floor is:

                     0.26337  

Step-by-step explanation:

We need to use the binomial theorem to find the probability.

The probability of k success in n experiments is given by:

       $P(X=k)=n_C_k\cdot p^k\cdot (1-p)^{n-k}$

where p is the probability of success.

Here p=1/3

( It represents the probability of choosing first floor)

k=1 ( since only one customer has to chose first floor)

n=6 since there are a total of 6 customers.

This means that:

$P(X=1)=6_C_1\times (\dfrac{1}{3})^1\times (1-\dfrac{1}{3})^{6-1}\\\\\\P(X=1)=6\times (\dfrac{1}{3})\times (\dfrac{2}{3})^5\\\\\\P(X=1)=0.26337$

Answer 2

Using the binomial distribution, it is found that there is a 0.2634 = 26.34% probability that exactly one customer dines on the first floor.

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For each customer, there are only two possible outcomes, either they dine on the first floor, or they do not. The probability of a customer dining on the first floor is independent of any other customer, which means that the binomial probability distribution is used to solve this question.

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Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of a success on a single trial.

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• Six customers, thus $n = 6$.
• They are equally as likely to dine on any of the three floors, thus $p = \frac{1}{3} = 0.3333$.

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The probability that exactly one customer dines on the first floor is P(X = 1), thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{6,1}.(0.3333)^{1}.(0.6667)^{5} = 0.2634$

0.2634 = 26.34% probability that exactly one customer dines on the first floor.

A similar problem is given at https://brainly.com/question/13036444