Question

Based on a​ poll, 66​% of Internet users are more careful about personal information when using a public​ Wi-Fi hotspot. What is the probability that among four randomly selected Internet​ users, at least one is more careful about personal information when using a public​ Wi-Fi hotspot? How is the result affected by the additional information that the survey subjects volunteered to​ respond? The probability that at least one of them is careful about personal information is nothing. ​(Round to three decimal places as​ needed.) How is the result affected by the additional information that the survey subjects volunteered to​ respond? A. It is very possible that the result is not valid because not everyone uses public​ Wi-Fi. B. It is very possible that the result is not valid because the sample may not be representative of the people who use public​ Wi-Fi. C. The result should not be impacted by this because volunteers are likely to have the most relevant responses. D. The result should not be impacted by this because there is no reason to lie when responding.

Answer 1

Answer:

The probability that at least 1 is more careful about personal information when using a public​ Wi-Fi hotspot is 0.987.

Correct option is (A).

Step-by-step explanation:

Let X = number of Internet users who are more careful about personal information when using a public​ Wi-Fi hotspot.

The probability of the random variable X is, p = 0.66.

A random sample of n = 4 internet users are selected.

The random variable X follows a Binomial distribution with parameters n = 4 and p = 0.66.

The probability mass function of X is:

$P(X=x)={4\choose x}0.66^{x}(1-0.66)^{4-x};\ x=0,1,2,3...$

Compute the probability that at least 1 is more careful about personal information when using a public​ Wi-Fi hotspot as follows:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             $=1-{4\choose 0}0.66^{0}(1-0.66)^{4-0}\\=1-(1\times 1\times 0.0134)\\=1-0.0134\\=0.9866\\\approx0.987$

Thus, the probability that at least 1 is more careful about personal information when using a public​ Wi-Fi hotspot is 0.987.

If the survey subjects volunteered to respond to the survey, then the sampling method used is Volunteering sampling. This type of sampling involves bias caused by the subjects known as voluntary response bias.

The bias occurs because the people could be responding uninterested in the survey or they could be emotionally driven to answer the survey.

This bias can affect the probability of the random variable, causing a huge probability of success.

Many people uses public wifi so the proportion of people more careful must be high.

So the correct option is (A).

Answer 2

Let X become a random variable representing the percentage of web users who are more cautious about private details when accessing a public WiFi hotspot in a group of four internet users.

• Consider that 66% of online consumers are much more cautious when accessing communal WiFi networks.
• Considering "identifying the internet user who is much more cautious about personal information when utilizing a wifi hotspot" to be effective.

               $\to \text{Probability of success (p)} = 0.66\\\\\to \text{Numuber of trials (n)} = 4$

• Since the probability of success remains constant in each trial, the number of trials is finite, the outcomes are independent, and there are only two mutually exclusive results (victory and inability) for every trial.
• we can assume that X follows a random variable with parameters $n = 4 \ \ and\ \ p = 0.66.$
• The binary probability law provides a likelihood of exactly x wins in n trials, which is as continues to follow:

              $\to \large P(X = x) = \binom{n}{x}(p)^{x}(1-p)^{n-x}$

• Where p is the probability of success.  We have to find $P(X = \text{at least} 1)$.

           $\to P(X =\ at \ least\ 1) = 1 - P(X = 0) \\\\\to \large P(X = at\ least\ 1) = 1 - \binom{4}{0}(0.66)^{0}(1-0.66)^{4-0}\\\\\to \large \left [ \because \binom{n}{0} = 1\ \ and\ \ (p)^{0} = 1 \right ]\\\\\to \large P(X = at\ least\ 1) = 1 - (0.34)^{4} \\\\\to \large P(X = at\ least\ 1) = 0.987$

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