To solve this problem, we need to find the probability that at most nine out of eleven babies are girls, given that each baby has an equal chance of being a boy or a girl (i.e., probability of being a girl is 0.5).
This situation can be modeled using a binomial distribution, as we are dealing with a fixed number of independent trials (babies) with two possible outcomes (girl or boy) for each trial. The binomial probability for exactly [tex]\( k \)[/tex] girls out of [tex]\( n \)[/tex] babies is given by:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
Where:
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient, which represents the number of ways to choose [tex]\( k \)[/tex] girls out of [tex]\( n \)[/tex] babies,
- [tex]\( n \)[/tex] is the total number of trials (babies),
- [tex]\( k \)[/tex] is the number of successes (girls),
- [tex]\( p \)[/tex] is the probability of success on an individual trial (0.5 in this case).
We need to find the probability of having at most nine girls, which means we need to sum the probabilities of having 0, 1, 2, ..., 9 girls.
Using the cumulative distribution function (CDF) of the binomial distribution, we can calculate this cumulative probability directly.
The probability that at most nine of the eleven babies are girls is approximately 0.994140625.
Given the choices:
- A. [tex]\(\frac{191}{256}\)[/tex] ≈ 0.74609375
- B. [tex]\(\frac{23}{250}\)[/tex] ≈ 0.092
- C. [tex]\(\frac{397}{2048}\)[/tex] ≈ 0.19384765625
- D. [tex]\(\frac{500}{512}\)[/tex] ≈ 0.9765625
Among these, the closest match to our calculated probability (0.994140625) is option D: [tex]\(\frac{500}{512}\)[/tex].
Therefore, the probability that at most nine out of eleven babies are girls is:
D. [tex]\(\frac{500}{512}\)[/tex]
