Find trusted answers to your questions with the help of IDNLearn.com's knowledgeable community. Ask anything and receive well-informed answers from our community of experienced professionals.
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 using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. IDNLearn.com is your source for precise answers. Thank you for visiting, and we look forward to helping you again soon.
