In a large city, 82% of residents own a cell phone. Suppose that we randomly select three city residents. What is the probability that at least one of the three residents does not own a cell phone

P(X≥1) = 0.45 (approximately)

Step-by-step explanation:

Let X be a random variable, then X follows a binomial distribution with probability of success, p and number trials, n.

The probability distribution is given as:

P(X) = nCx * p^x * (1 – p)^(n – x)

Where p is Probability of success, n is number of trials and x is total number of successes.

Probability of having phone is 82% = 0.82

Probability of not having phone is 100% – 82% = 0.18.

Since the trials are interested in those that are not having phone, then probability of success, p = 0.18.

Number of trials, n = 3.

P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0)

P(X = 0) = 3C0 * (0.18)^0 * (0.82)^(3) = 1 * 1 * 0.551368 = 0.551368

P(X ≥ 1) = 1 – 0.551368 = 0.448632 ≈ 0.45

i used to know how to do this in 7th but i forget easily

step-by-step explanation: