Question

One day, thirteen babies are born at a hospital. assuming each baby has an equal chance of being a boy or girl, what is the probability that at most eleven of the thirteen babies are girls?

Answer 1

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:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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:

$P(X \leq 11) = 1 - P(X > 11)$

In which

P(X > 11) = P(X = 12) + P(X = 13)

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 12) = C_{13,12}.(0.5)^{12}.(0.5)^{1} = 0.0016$

$P(X = 13) = C_{13,13}.(0.5)^{13}.(0.5)^{0} = 0.0001$

So:

P(X > 11) = P(X = 12) + P(X = 13) = 0.0016 + 0.0001 = 0.0017

$P(X \leq 11) = 1 - P(X > 11) = 1 - 0.0017 = 0.9983$

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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