Final answer:
The probability that at least one of them is a girl is 0.988.
Step-by-step explanation:
To solve this problem, we'll use the concept of complementary probability. The complementary probability of the event "at least one of them is a girl" is the event "all of them are boys".
Let's denote:
- P(boy) as the probability of a baby being a boy, which is 0.526.
- P(girl) as the probability of a baby being a girl, which is 1 - P(boy) because there are only two possible outcomes (boy or girl) and the probabilities must add up to 1.
Therefore, P(girl) = 1 - 0.526 = 0.474.
We want to calculate P(at least one girl in nine births), which we can obtain by subtracting the probability of its complement (which is P(all boys in nine births)) from 1.
The probability of all nine births being boys is P(all boys) = (P(boy))^9.
So, P(all boys) = 0.526^9.
Now, let's compute this:
P(all boys) = 0.526^9 ≈ 0.011988 (rounded to six decimal places for intermediate step accuracy)
Next, we calculate the probability of at least one girl by taking the complement of P(all boys):
P(at least one girl) = 1 - P(all boys)
P(at least one girl) = 1 - 0.011988
P(at least one girl) ≈ 0.988012
Now, let's round this to three decimal places as instructed:
P(at least one girl) ≈ 0.988
So, the probability that among the next nine randomly selected births in the country, at least one of them is a girl, is approximately 0.988 when rounded to three decimal places.