Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Answer:
0.1143 = 11.43% probability that all but one of them are using Chrome as their browser
Step-by-step explanation:
For each person, there are only two possible outcomes. Either they use Chrome, or they do not. The probability of a person using Chrome is independent of any other person, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Chrome: 52.57%
This means that [tex]p = 0.5257[/tex]
Sample of 6 people
This means that [tex]n = 6[/tex]
What is the probability that all but one of them are using Chrome as their browser?
5 using Chrome, so:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{6,5}.(0.5257)^{5}.(1-0.5257)^{1} = 0.1143[/tex]
0.1143 = 11.43% probability that all but one of them are using Chrome as their browser
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
