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Sagot :
To determine whether events [tex]\( A \)[/tex] (a student is in eleventh grade) and [tex]\( B \)[/tex] (a student is enrolled in French class) are independent, we need to verify if the conditional probability [tex]\( P(A \mid B) \)[/tex] is equal to the probability [tex]\( P(A) \)[/tex].
1. Calculate [tex]\( P(A) \)[/tex]: Probability that a student is in eleventh grade
The total number of students is 552. The number of eleventh-grade students is 138.
[tex]\[ P(A) = \frac{\text{Number of eleventh-grade students}}{\text{Total number of students}} = \frac{138}{552} \][/tex]
2. Calculate [tex]\( P(B) \)[/tex]: Probability that a student is enrolled in French class
The number of students enrolled in the French class is 272.
[tex]\[ P(B) = \frac{\text{Number of students enrolled in French class}}{\text{Total number of students}} = \frac{272}{552} \][/tex]
3. Calculate [tex]\( P(A \cap B) \)[/tex]: Probability that a student is in eleventh grade and enrolled in French class
The number of eleventh-grade students enrolled in the French class is 68.
[tex]\[ P(A \cap B) = \frac{\text{Number of eleventh-grade students enrolled in French class}}{\text{Total number of students}} = \frac{68}{552} \][/tex]
4. Calculate [tex]\( P(A \mid B) \)[/tex]: Probability that a student is in eleventh grade given that they are enrolled in French class
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{68}{552}}{\frac{272}{552}} = \frac{68}{272} \][/tex]
Simplify the fraction:
[tex]\[ P(A \mid B) = \frac{68}{272} = \frac{1}{4} \][/tex]
5. Check if [tex]\( P(A \mid B) \)[/tex] is equal to [tex]\( P(A) \)[/tex]
We need to compare:
[tex]\[ P(A) = \frac{138}{552} = \frac{1}{4} \][/tex]
Since [tex]\( P(A \mid B) = P(A) \)[/tex],
[tex]\[ P(A \mid B) = \frac{1}{4} \text{ and } P(A) = \frac{1}{4} \][/tex]
[tex]\( P(A \mid B) \)[/tex] is indeed equal to [tex]\( P(A) \)[/tex].
Therefore, the correct statement is:
- [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are independent events because [tex]\( P(A \mid B) = P(A) \)[/tex].
So, the true statement about [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is:
[tex]\[ \boxed{A \text{ and } B \text{ are independent events because } P(A \mid B) = P(A).} \][/tex]
1. Calculate [tex]\( P(A) \)[/tex]: Probability that a student is in eleventh grade
The total number of students is 552. The number of eleventh-grade students is 138.
[tex]\[ P(A) = \frac{\text{Number of eleventh-grade students}}{\text{Total number of students}} = \frac{138}{552} \][/tex]
2. Calculate [tex]\( P(B) \)[/tex]: Probability that a student is enrolled in French class
The number of students enrolled in the French class is 272.
[tex]\[ P(B) = \frac{\text{Number of students enrolled in French class}}{\text{Total number of students}} = \frac{272}{552} \][/tex]
3. Calculate [tex]\( P(A \cap B) \)[/tex]: Probability that a student is in eleventh grade and enrolled in French class
The number of eleventh-grade students enrolled in the French class is 68.
[tex]\[ P(A \cap B) = \frac{\text{Number of eleventh-grade students enrolled in French class}}{\text{Total number of students}} = \frac{68}{552} \][/tex]
4. Calculate [tex]\( P(A \mid B) \)[/tex]: Probability that a student is in eleventh grade given that they are enrolled in French class
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{68}{552}}{\frac{272}{552}} = \frac{68}{272} \][/tex]
Simplify the fraction:
[tex]\[ P(A \mid B) = \frac{68}{272} = \frac{1}{4} \][/tex]
5. Check if [tex]\( P(A \mid B) \)[/tex] is equal to [tex]\( P(A) \)[/tex]
We need to compare:
[tex]\[ P(A) = \frac{138}{552} = \frac{1}{4} \][/tex]
Since [tex]\( P(A \mid B) = P(A) \)[/tex],
[tex]\[ P(A \mid B) = \frac{1}{4} \text{ and } P(A) = \frac{1}{4} \][/tex]
[tex]\( P(A \mid B) \)[/tex] is indeed equal to [tex]\( P(A) \)[/tex].
Therefore, the correct statement is:
- [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are independent events because [tex]\( P(A \mid B) = P(A) \)[/tex].
So, the true statement about [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is:
[tex]\[ \boxed{A \text{ and } B \text{ are independent events because } P(A \mid B) = P(A).} \][/tex]
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