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Sagot :
Using the binomial distribution, it is found that there is a 0.999977 = 99.9977% probability that at least one of them makes a Type I error.
What is the binomial distribution formula?
The formula is:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- There are 25 students, hence n = 25.
- 70% do not commit any error, hence 30% do and p = 0.3.
The probability that at least one commits an error is given by:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{30,0}.(0.3)^{0}.(0.7)^{30} = 0.000023[/tex]
Then:
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.000023 = 0.999977[/tex]
There is a 0.999977 = 99.9977% probability that at least one of them makes a Type I error.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
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