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Belleville High School offers classes in three different foreign languages. Let [tex]$A$[/tex] be the event that a student is in eleventh grade, and let [tex]$B$[/tex] be the event that a student is enrolled in a French class.

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
& Spanish & French & German & Total \\
\hline
Tenth Grade & 107 & 122 & 6 & 235 \\
\hline
Eleventh Grade & 56 & 68 & 14 & 138 \\
\hline
Twelfth Grade & 89 & 82 & 8 & 179 \\
\hline
Total & 262 & 272 & 28 & 552 \\
\hline
\end{tabular}
\][/tex]

Which statement is true about whether [tex]$A$[/tex] and [tex]$B$[/tex] are independent events?

A. [tex]$A$[/tex] and [tex]$B$[/tex] are independent events because [tex]$P(A \mid B) = P(A)$[/tex].

B. [tex]$A$[/tex] and [tex]$B$[/tex] are independent events because [tex]$P(A \mid B) = P(B)$[/tex].

C. [tex]$A$[/tex] and [tex]$B$[/tex] are not independent events because [tex]$P(A \mid B) \neq P(A)$[/tex].

D. [tex]$A$[/tex] and [tex]$B$[/tex] are not independent events because [tex]$P(A \mid B) \neq P(B)$[/tex].


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]