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

a precentage of 6th graders. the population is all 6th graders. let me know if i'm wrong, i hope this ! : )

step-by-step explanation: