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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
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