Question

According to a 2018 survey, 74 percent of employed young adults expect to bring work on a vacation trip. A random sample of 20 employed young adults will be selected. What is the probability that 8 of the selected young adults will expect to bring work on a vacation trip?

Answer 1

Answer:

$P(X=8)=(20C8)(0.74)^8 (1-0.74)^{20-8}=0.00108$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we can assume that follows this distribution:

$X \sim Binom(n=20, p=0.74)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

Solution to the problem

For this case we want this probability:

$P(X=8)$

And if we replace into the probability mass function we got:

$P(X=8)=(20C8)(0.74)^8 (1-0.74)^{20-8}=0.00108$