Question

belleville high school offers classes on three different foreign languages. let a be the event that a student is in eleventh grade, and let b be the event that a student is enrolled in french class.

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$$\begin{tabular}{|c|c|c|c|c|} \\hline & spanish & french & german & total \\\\ \\hline tenth grade & 107 & 122 & 6 & 235 \\\\ \\hline eleventh grade & 56 & 68 & 14 & 138 \\\\ \\hline twelfth grade & 89 & 82 & 8 & 179 \\\\ \\hline total & 252 & 272 & 28 & 552 \\\\ \\hline \\end{tabular}$$

which statement is true about whether a and b are independent events?

\\(a\\) and \\(b\\) are independent events because \\(p(a|b) = p(a)\\).
\\(a\\) and \\(b\\) are independent events because \\(p(a|b) = p(b)\\).
\\(a\\) and \\(b\\) are not independent events because \\(p(a|b) \
eq p(a)\\).
\\(a\\) and \\(b\\) are not independent events because \\(p(a|b) \
eq p(b)\\).

Explanation:

Define the events and extract values

Using the Two-Way Frequency Tables knowledge point

• Total number of students: \(N = 552\)
• Event \(A\) (Eleventh Grade): \(n(A) = 138\)
• Event \(B\) (French Class): \(n(B) = 272\)
• Intersection \(A \cap B\) (Eleventh Grade and French Class): \(n(A \cap B) = 68\)

Calculate individual and conditional probabilities

Using the Conditional Probability Calculation knowledge point

$$ P(A) = \frac{138}{552} = 0.25 $$

$$ P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{68}{272} = 0.25 $$

Determine independence of events

Using the Independent Events knowledge point

$$ P(A|B) = P(A) = 0.25 $$

Since the conditional probability of \(A\) given \(B\) is equal to the marginal probability of \(A\), the events \(A\) and \(B\) are independent.

Answer:

• A and B are independent events because P(A|B) = P(A). (Correct answer)
• A and B are independent events because P(A|B) = P(B).
• A and B are not independent events because P(A|B) ≠ P(A).
• A and B are not independent events because P(A|B) ≠ P(B).