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Sagot :
Sure, let's analyze the problem step-by-step:
1. Identify the Sample Space [tex]\( S \)[/tex]:
We need to find the set of outcomes for the possible genders of two children. If [tex]\( B \)[/tex] represents a boy and [tex]\( G \)[/tex] represents a girl, then the sample space is:
[tex]\[ S = \{BB, BG, GB, GG\} \][/tex]
where the first letter represents the gender of the oldest child and the second letter represents the gender of the youngest child.
2. Define the Random Variable [tex]\( X \)[/tex]:
Let [tex]\( X \)[/tex] represent the number of times [tex]\( G \)[/tex] (a girl) occurs in each outcome. The possible values of [tex]\( X \)[/tex] are [tex]\( 0, 1, \)[/tex] and [tex]\( 2 \)[/tex]:
- [tex]\( X = 0 \)[/tex] when there are no girls (outcome: [tex]\( BB \)[/tex]).
- [tex]\( X = 1 \)[/tex] when there is exactly one girl (outcomes: [tex]\( BG \)[/tex] and [tex]\( GB \)[/tex]).
- [tex]\( X = 2 \)[/tex] when there are two girls (outcome: [tex]\( GG \)[/tex]).
3. Count the Frequencies of Each [tex]\( X \)[/tex] Value:
We now count how often each value of [tex]\( X \)[/tex] occurs in the sample space [tex]\( S \)[/tex]:
- [tex]\( X = 0 \)[/tex]: Only one outcome, [tex]\( BB \)[/tex], fits this criterion.
- [tex]\( X = 1 \)[/tex]: Two outcomes, [tex]\( BG \)[/tex] and [tex]\( GB \)[/tex], each fit this criterion.
- [tex]\( X = 2 \)[/tex]: Only one outcome, [tex]\( GG \)[/tex], fits this criterion.
4. Calculate the Probabilities:
Each outcome in the sample space [tex]\( S \)[/tex] is equally probable. Since there are 4 outcomes, the probability of each outcome is [tex]\( \frac{1}{4} \)[/tex].
- Probability that [tex]\( X = 0 \)[/tex]: There is 1 outcome out of 4 where there are no girls.
[tex]\[ P_X(0) = \frac{1}{4} = 0.25 \][/tex]
- Probability that [tex]\( X = 1 \)[/tex]: There are 2 outcomes out of 4 where there is exactly one girl.
[tex]\[ P_X(1) = \frac{2}{4} = 0.5 \][/tex]
- Probability that [tex]\( X = 2 \)[/tex]: There is 1 outcome out of 4 where there are two girls.
[tex]\[ P_X(2) = \frac{1}{4} = 0.25 \][/tex]
5. Conclusion:
The probability distribution [tex]\( P_X(x) \)[/tex] for the random variable [tex]\( X \)[/tex] (the number of girls) is as follows:
[tex]\[ \begin{tabular}{|c|c|} \hline X & P_X(x) \\ \hline 0 & 0.25 \\ \hline 1 & 0.5 \\ \hline 2 & 0.25 \\ \hline \end{tabular} \][/tex]
Given the choices in the problem statement, this matches the first table provided:
[tex]\[ \begin{tabular}{|c|c|} \hline X & P_X(x) \\ \hline 0 & 0.25 \\ \hline 1 & 0.5 \\ \hline 2 & 0.25 \\ \hline \end{tabular} \][/tex]
So, the correct probability distribution is the one in the first table.
1. Identify the Sample Space [tex]\( S \)[/tex]:
We need to find the set of outcomes for the possible genders of two children. If [tex]\( B \)[/tex] represents a boy and [tex]\( G \)[/tex] represents a girl, then the sample space is:
[tex]\[ S = \{BB, BG, GB, GG\} \][/tex]
where the first letter represents the gender of the oldest child and the second letter represents the gender of the youngest child.
2. Define the Random Variable [tex]\( X \)[/tex]:
Let [tex]\( X \)[/tex] represent the number of times [tex]\( G \)[/tex] (a girl) occurs in each outcome. The possible values of [tex]\( X \)[/tex] are [tex]\( 0, 1, \)[/tex] and [tex]\( 2 \)[/tex]:
- [tex]\( X = 0 \)[/tex] when there are no girls (outcome: [tex]\( BB \)[/tex]).
- [tex]\( X = 1 \)[/tex] when there is exactly one girl (outcomes: [tex]\( BG \)[/tex] and [tex]\( GB \)[/tex]).
- [tex]\( X = 2 \)[/tex] when there are two girls (outcome: [tex]\( GG \)[/tex]).
3. Count the Frequencies of Each [tex]\( X \)[/tex] Value:
We now count how often each value of [tex]\( X \)[/tex] occurs in the sample space [tex]\( S \)[/tex]:
- [tex]\( X = 0 \)[/tex]: Only one outcome, [tex]\( BB \)[/tex], fits this criterion.
- [tex]\( X = 1 \)[/tex]: Two outcomes, [tex]\( BG \)[/tex] and [tex]\( GB \)[/tex], each fit this criterion.
- [tex]\( X = 2 \)[/tex]: Only one outcome, [tex]\( GG \)[/tex], fits this criterion.
4. Calculate the Probabilities:
Each outcome in the sample space [tex]\( S \)[/tex] is equally probable. Since there are 4 outcomes, the probability of each outcome is [tex]\( \frac{1}{4} \)[/tex].
- Probability that [tex]\( X = 0 \)[/tex]: There is 1 outcome out of 4 where there are no girls.
[tex]\[ P_X(0) = \frac{1}{4} = 0.25 \][/tex]
- Probability that [tex]\( X = 1 \)[/tex]: There are 2 outcomes out of 4 where there is exactly one girl.
[tex]\[ P_X(1) = \frac{2}{4} = 0.5 \][/tex]
- Probability that [tex]\( X = 2 \)[/tex]: There is 1 outcome out of 4 where there are two girls.
[tex]\[ P_X(2) = \frac{1}{4} = 0.25 \][/tex]
5. Conclusion:
The probability distribution [tex]\( P_X(x) \)[/tex] for the random variable [tex]\( X \)[/tex] (the number of girls) is as follows:
[tex]\[ \begin{tabular}{|c|c|} \hline X & P_X(x) \\ \hline 0 & 0.25 \\ \hline 1 & 0.5 \\ \hline 2 & 0.25 \\ \hline \end{tabular} \][/tex]
Given the choices in the problem statement, this matches the first table provided:
[tex]\[ \begin{tabular}{|c|c|} \hline X & P_X(x) \\ \hline 0 & 0.25 \\ \hline 1 & 0.5 \\ \hline 2 & 0.25 \\ \hline \end{tabular} \][/tex]
So, the correct probability distribution is the one in the first table.
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