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Sagot :
Using the binomial distribution, there is a 0.9983 = 99.83% probability that at most eleven of the thirteen babies are girls.
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.
For this problem, the values of the parameters are:
p = 0.5, n = 13
The probability that at most eleven of the thirteen babies are girls is:
[tex]P(X \leq 11) = 1 - P(X > 11)[/tex]
In which
P(X > 11) = P(X = 12) + P(X = 13)
Then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 12) = C_{13,12}.(0.5)^{12}.(0.5)^{1} = 0.0016[/tex]
[tex]P(X = 13) = C_{13,13}.(0.5)^{13}.(0.5)^{0} = 0.0001[/tex]
So:
P(X > 11) = P(X = 12) + P(X = 13) = 0.0016 + 0.0001 = 0.0017
[tex]P(X \leq 11) = 1 - P(X > 11) = 1 - 0.0017 = 0.9983[/tex]
0.9983 = 99.83% probability that at most eleven of the thirteen babies are girls.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
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