Question

When Ryan is serving at a restaurant, there is a 0.75 probability that each party will order drinks with their meal. During one hour, Ryan served 6 parties. Assuming that each party is equally likely to order drinks, what is the probability that at least one party will not order drinks? Round your answer to the nearest hundredth.

Answer 1

Answer:

0.82 = 82% probability that at least one party will not order drinks

Step-by-step explanation:

For each party, there are only two possible outcomes. Either they will order drinks with their meal, or they will not. The probability of a party ordering drinks with their meal is independent of other parties. So the binomial probability distribution is used to solve this question.

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 X happening.

When Ryan is serving at a restaurant, there is a 0.75 probability that each party will order drinks with their meal.

This means that $p = 0.75$

During one hour, Ryan served 6 parties. Assuming that each party is equally likely to order drinks, what is the probability that at least one party will not order drinks?

6 parties, so n = 6.

Either all parties will order drinks, or at least one will not. The sum of the probabilities of these events is decimal 1. So

$P(X = 6) + P(X < 6) = 1$

We want P(X < 6). So

$P(X < 6) = 1 - P(X = 6)$

In which

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

$P(X = 6) = C_{6,6}.(0.75)^{6}.(0.25)^{0} = 0.18$

$P(X < 6) = 1 - P(X = 6) = 1 - 0.18 = 0.82$

0.82 = 82% probability that at least one party will not order drinks

Answer 2

Answer:

the answer  that you are looking for is 82 percent

Step-by-step explanation: