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Sagot :
Using the binomial distribution, it is found that there is a 0.7447 = 74.47% probability that fewer than 7 of them show up.
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, the values of the parameters are given as follows:
p = 0.7, n = 8.
The probability that fewer than 7 of them show up is given by:
[tec]P(X < 7) = 1 - P(X \geq 7)[/tex]
In which:
[tex]P(X \geq 7) = P(X = 7) + P(X = 8)[/tex]
Then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 7) = C_{8,7}.(0.7)^{7}.(0.3)^{1} = 0.1977[/tex]
[tex]P(X = 8) = C_{8,8}.(0.7)^{8}.(0.3)^{0} = 0.0576[/tex]
Then:
[tex]P(X \geq 7) = P(X = 7) + P(X = 8) = 0.1977 + 0.0576 = 0.2553[/tex]
[tec]P(X < 7) = 1 - P(X \geq 7) = 1 - 0.2553 = 0.7447[/tex]
0.7447 = 74.47% probability that fewer than 7 of them show up.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
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