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Chapter 5: Problem 60

According to the National Center for Health Statistics, \(52 \%\) of U.S. households no longer have a landline and instead only have cell phone service. Suppose three U.S. households are selected at random. a. What is the probability that all three have only cell phone service? b. What is the probability that at least one has only cell phone service?

Short Answer

Expert verified
a) The probability that all three households have only cell phone service is approximately 14.06%. b) The probability that at least one household has only cell phone service is approximately 88.94%.

Step by step solution

01

Identify Necessary Values for the Distribution

The probability of a household having only cell phone service is \(0.52\), or \(52\%\). There are \(n=3\) trials (3 households are selected). The binomial distribution comes into play because we are repeating a 'Bernoulli trial' (cell phone only or not) multiple times.
02

Calculate the Probability for Case A

In case a, we want the probability that all three households have only cell phone service. In a binomial distribution, the probability of exactly k successes (k = 3 in this case) in n trials (n = 3 in this case) is given by \(P(K=k) = C(n, k) \cdot p^k \cdot (1-p)^{n-k}\). Here, \(p = 0.52\). So, we substitute these values into the equation, and we get \(P(K=3) = C(3, 3) \cdot (0.52)^3 \cdot (1-0.52)^{3-3} = 1 \cdot 0.140608 \cdot 1 = 0.140608, or 14.06\%\).
03

Calculate the Probability for Case B

In case b, we want the probability that at least one household has only cell phone service. This is the complement of the event that none of the three has cell phone service only. Therefore, we can calculate the probability that none of the households has cell phone service only, and subtract this from 1. So using the formula from Step 2, we calculate the probability for k = 0. \(P(K=0) = C(3, 0) \cdot (0.52)^0 \cdot (1-0.52)^{3-0} = 1 \cdot 1 \cdot 0.11059168 = 0.11059168, or 11.06\%\). Therefore, the probability that at least one household having cell phone service only, \(P(K \geq 1) = 1 - P(K=0) = 1 - 0.11059168 = 0.88940832, or 88.94\%).

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