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Aus Erfahrungswerten sei bekannt, dab ein neugeborenes Kind mit WahrscheinLichkeit 0,515 ein Junge ist. Wie groß ist die Wahrscheinlichkeit dafur, dab in einer Familie mit sechs Kindern a) alle Kinder Mädchen, b) wenigstens 5 der Kinder Mädchen, c) wenigstens 3 der Kinder Mädchen sind?

Short Answer

Expert verified
a) 0.0114, b) 0.0748, c) 0.5611

Step by step solution

01

Title – Define Key Probabilities

Let the probability of a child being a boy be denoted as \( P(B) = 0.515 \). Therefore, the probability of a child being a girl is \( P(G) = 1 - P(B) = 1 - 0.515 = 0.485 \).
02

Title – Define the Binomial Distribution

The number of girls in a family of six children follows a binomial distribution: \[ P(X=k) = \binom{6}{k} (0.485)^k (0.515)^{6-k} \] where \( X \) is the number of girls and \( k \) is the specific number of girls.
03

Title – Calculate Probability for All Girls (a)

To find the probability that all six children are girls, we set \( k = 6 \): \[ P(X=6) = \binom{6}{6} (0.485)^6 (0.515)^0 = 1 \cdot 0.485^6 \approx 0.0114 \]
04

Title – Calculate Probability for At Least 5 Girls (b)

To find the probability that at least 5 of the six children are girls, sum up the probabilities for 5 and 6 girls: \[ P(X \geq 5) = P(X=5) + P(X=6) \] Calculate \( P(X=5) \): \[ P(X=5) = \binom{6}{5} (0.485)^5 (0.515)^1 = 6 \cdot 0.485^5 \cdot 0.515 \approx 0.0634 \] Then, \[ P(X \geq 5) = 0.0634 + 0.0114 = 0.0748 \]
05

Title – Calculate Probability for At Least 3 Girls (c)

To find the probability that at least 3 of the six children are girls, sum the probabilities for 3, 4, 5, and 6 girls: \[ P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) + P(X=6) \] Calculate each: \[ P(X=3) = \binom{6}{3} (0.485)^3 (0.515)^3 = 20 \cdot 0.485^3 \cdot 0.515^3 \approx 0.2781 \] \[ P(X=4) = \binom{6}{4} (0.485)^4 (0.515)^2 = 15 \cdot 0.485^4 \cdot 0.515^2 \approx 0.2082 \] \[ P(X=5) = 0.0634 \] \[ P(X=6) = 0.0114 \] Then, \[ P(X \geq 3) = 0.2781 + 0.2082 + 0.0634 + 0.0114 = 0.5611 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

binomial distribution
A binomial distribution is a discrete probability distribution. It describes the number of successes in a fixed number of independent experiments. Each experiment, or trial, has two possible outcomes: success or failure.
In our exercise, the success is defined as a child being a girl, and the number of trials is six, since the family has six children. The probability of success (a girl being born) is 0.485, and the probability of failure (a boy being born) is 0.515.
The binomial distribution formula we use looks like this:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Here, \( n \) is the total number of trials, \( k \) is the number of successes, \( p \) is the probability of success, and \( 1-p \) is the probability of failure.
Using this formula, we can compute the probability for any number of girls in the family.
probability calculation
Probability calculation involves determining the likelihood of specific outcomes. To find the probability of certain numbers of girls in a family using the binomial distribution, we use the combination formula and the success/failure probabilities.
For example, calculating for all girls (no boys) in the family of six:
\[ P(X=6) = \binom{6}{6} (0.485)^6 (0.515)^0 \]
This formula translates to:
\[ 1 \cdot 0.485^6 = 0.0114 \]
Similarly, we can follow the same steps for other cases, such as for at least 5 girls. We sum up the probabilities for having exactly 5 girls and exactly 6 girls:
\[ P(X \geq 5) = P(X=5) + P(X=6) \]
Each probability calculation builds on this structure.
combinatorial analysis
Combinatorial analysis is a key part of calculating binomial probabilities. It involves the use of combinations, which are selections of items from a larger set without considering the order.
The combination formula is:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
In the context of our exercise, this formula helps to determine the number of ways to choose \( k \) girls out of \( n \) children. For instance, if we need the number of ways to have 5 girls out of 6 children, it's calculated as:
\[ \binom{6}{5} = \frac{6!}{5!(6-5)!} = 6 \]
Combinatorial terms are then multiplied by the success and failure probabilities to get binomial probabilities.
Without combinatorial analysis, determining these probabilities for various outcomes would be difficult.
probability theory
Probability theory underpins all our calculations through the concepts of random experiments, outcomes, and events.
A core principle is that the sum of probabilities of all possible outcomes must equal 1. For example, in our problem, we're assured that:
\[ P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) = 1 \]
Probability theory also ensures that we can calculate complement events easily. For instance, to find the probability of 'at least 3 girls', we can sum up the probabilities from 3 to 6 directly, as shown:
\[ P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) + P(X=6) \]
It simplifies complex probability calculations, making it easier to deal with real-world scenarios.

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