Question

According to the National Telecommunication and Information Administration, 51% of U.S. households had Internet access in 2001.

a. If we select four U.S. households at random (with replacement), what is the probability that all four had Internet access in 2001?
b. What is the probability that at least one of four randomly selected U.S. households had Internet access in 2001?

Answer 1

a

`P(X = 4 ) = 0.0677`

b

`P(X \ge 1  ) =   0.9424`

Explanation:

From the question we are told that

The probability of a US household having an internet access in 2001 is p = 0.51

The sample size is n = 4

Generally the distribution of the number of US household with internet access at 2001 follows a binomial distribution

i.e

`X  \~ \ \ \  B(n , p)`

and the probability distribution function for binomial distribution is

`P(X = x) =  ^(n)C_x *  p^x *  (1- p)^(n-x)`

Here C stands for combination hence we are going to be making use of the combination function in our calculators

Generally the probability that all four had Internet access in 2001 is mathematically represented as

`P(X = 4 ) =  ^(4)C_4 *  (0.51)^4  (1- 0.51)^(4-4)`

=>
`P(X = 4 ) = 0.0677`

Generally the probability that at least one of four randomly selected U.S. households had Internet access in 2001 is mathematically represented as

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

=>
`P(X \ge 1  ) =   1-  [^(4)C_0 *  (0.51 )^0 *  (1- 0.51 )^(4-0 ) ]`

=>
`P(X \ge 1  ) =   1-  [1 *  1 *  (0.49 )^(4 ) ]`

=>
`P(X \ge 1  ) =   0.9424`