Question

according to a survey conducted by a credit card company, 71% of adults stated they want access to their cell phones and laptops while on vacation. if six adults are selected at random, determine the probability that exactly three of them want access to their cell phones and laptops while on vacation. the probability is (type an integer or decimal rounded to five decimal places as needed.)

Explanation:

Step1: Identify the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Determine the values of \(n\), \(k\), and \(p\)

Here, \(n = 6\) (number of adults selected), \(k=3\) (number of adults who want access), and \(p = 0.71\) (probability that an adult wants access). Then \(1-p=1 - 0.71=0.29\)

Step3: Calculate the combination \(C(n,k)\)

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Step4: Calculate \(p^{k}\) and \((1 - p)^{n - k}\)

\(p^{k}=(0.71)^{3}=0.71\times0.71\times0.71 = 0.357911\)
\((1 - p)^{n - k}=(0.29)^{3}=0.29\times0.29\times0.29=0.024389\)

Step5: Calculate the probability \(P(X = 3)\)

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Answer:

\(0.17454\)