Question

Suppose that 50% of all babies born in a particular hospital are boys. If 6 babies born in the hospital are randomly selected, what is the probability that fewer than 3 of them are boys?

Answer 1

Using the binomial distribution, it is found that there is a 0.3438 = 34.38% probability that fewer than 3 of them are boys.

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.

In this problem, the values of the parameters are given as follows:

n = 6, p = 0.5.

The probability that fewer than 3 of them are boys is given by:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which:

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

$P(X = 0) = C_{6,0}.(0.5)^{0}.(0.5)^{6} = 0.0156$

$P(X = 1) = C_{6,1}.(0.5)^{1}.(0.5)^{5} = 0.0938$

$P(X = 2) = C_{6,2}.(0.5)^{2}.(0.5)^{4} = 0.2344$

Then:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0156 + 0.0938 + 0.2344 = 0.3438$

0.3438 = 34.38% probability that fewer than 3 of them are boys.

More can be learned about the binomial distribution at brainly.com/question/24863377

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