According to the National Telecommunication and Information Administration, 56.5% of U. S. households owned a computer in 2001. What is the probability that of three randomly selected U. S. households at least one owned a computer in 2001?

Step-by-step explanation:

Assuming a binomial distribution for the number of U. S. households that owned a computer in 2001. The formula for binomial distribution is expressed as

P (X = r) = nCr * q^ (n - r) * p^r

Where

p = probability of success

q = probability of failure

n = number of sample

From the information given,

p = 56.5% = 56.5/190 = 0.565

q = 1 - p = 1 - 0.565 = 0.435

n = 3

We want to determine P (x greater than or equal to 1). This is also expressed as

1 - P (x lesser than or equal to 1)

P (x lesser than or equal to 1) = P (x = 0) + P (x = 1)

P (x = 0) = 3C0 * 0.435^ (3 - 0) * 0.565^0 = 0.082

P (x = 1) = 3C1 * 0.435^ (3 - 1) * 0.565^1 = 0.32

P (x greater than or equal to 1) = 0.082 + 0.32 = 0.402