Questions: In a certain country, the true probability of a baby being a boy is 0.539. Among the next four randomly selected births in the country, what is the probability that at least one of them is a girl? The probability is (Round to three decimal places as needed.)

Solution Steps

We are tasked with finding the probability that at least one of the next four randomly selected births in a country, where the probability of a baby being a boy is \( p = 0.539 \), is a girl. The probability of a baby being a girl is given by \( q = 1 - p = 0.461 \).

To find the probability of having at least one girl, we first calculate the probability of having no girls (i.e., all boys) in four births. This can be modeled using the binomial distribution:

\[ P(X = 0) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where:

• \( n = 4 \) (the number of births),
• \( x = 0 \) (the number of girls),
• \( p = 0.461 \) (the probability of a girl),
• \( q = 0.539 \) (the probability of a boy).

Calculating this gives:

\[ P(X = 0) = \binom{4}{0} \cdot (0.461)^0 \cdot (0.539)^4 = 1 \cdot 1 \cdot (0.539)^4 \approx 0.0844 \]

The probability of having at least one girl is the complement of the probability of having no girls:

\[ P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0844 \approx 0.916 \]

Final Answer