Question

According to a 2013 study by the Pew Research Center, 15% of adults in the United States do not use the Internet (Pew Research Center website, December, 15, 2014). Suppose that 10 adults in the United States are selected randomly.

A. What is the probability that all of the adults use the Internet (to 4 decimals)?

B. What is the probability that 3 of the adults do not use the Internet (to 4 decimals)? If you calculate the binomial probabilities manually, make sure to carry at least 4 decimal digits in your calculations.

C. What is the probability that at least 1 of the adults do not use the Internet (to 4 decimals)?

Answer 1

a)
`P(X=0)=(10C0) (0.15)^0 (1-0.15)^(10-0)= 0.1969`

b)
`P(X=3)=(10C3) (0.15)^3 (1-0.15)^(10-3)= 0.1298`

c)
`P(X \geq 1) = 1- P(X<1) = 1-P(X=0)`

`P(X=0)=(10C0) (0.15)^0 (1-0.15)^(10-0)= 0.1969`

So then we have:

`P(X \geq 1) = 1- P(X<1) = 1-P(X=0)= 1-0.1969= 0.8031`

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".

Solution to the problem

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=10, p=0.15)`

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=(n!)/((n-x)! x!)`

Part a

For this case we want this probability:

`P(X=0)=(10C0) (0.15)^0 (1-0.15)^(10-0)= 0.1969`

Part b

For this case we want this probability:

`P(X=3)`

And using the probability mass function we got:

`P(X=3)=(10C3) (0.15)^3 (1-0.15)^(10-3)= 0.1298`

Part c

For this case we want this probability:

`P(X \geq 1)`

And we can use the complenet rule and we got:

`P(X \geq 1) = 1- P(X<1) = 1-P(X=0)`

`P(X=0)=(10C0) (0.15)^0 (1-0.15)^(10-0)= 0.1969`

So then we have:

`P(X \geq 1) = 1- P(X<1) = 1-P(X=0)= 1-0.1969= 0.8031`