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. chart 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:

Step1: Calculate total number of students

Total students = 552 (from the "Total" row of the "Total" column).

Step2: Calculate P(A) (probability of being in eleventh grade)

Number of eleventh - grade students = 138.
So, \( P(A)=\frac{\text{Number of eleventh - grade students}}{\text{Total number of students}}=\frac{138}{552}=\frac{138\div138}{552\div138}=\frac{1}{4} = 0.25\)

Step3: Calculate P(B) (probability of being in French class)

Number of students in French class = 272.
So, \( P(B)=\frac{\text{Number of students in French class}}{\text{Total number of students}}=\frac{272}{552}=\frac{272\div8}{552\div8}=\frac{34}{69}\approx0.493\)

Step4: Calculate number of students in both eleventh grade and French class

From the table, number of eleventh - grade students in French class = 68.

Step5: Calculate P(A and B) (probability of being in eleventh grade and French class)

\( P(A\cap B)=\frac{\text{Number of students in both eleventh grade and French class}}{\text{Total number of students}}=\frac{68}{552}=\frac{68\div4}{552\div4}=\frac{17}{138}\approx0.123\)

Step6: Calculate P(A|B) (probability of being in eleventh grade given in French class)

By the formula \( P(A|B)=\frac{P(A\cap B)}{P(B)}\)
We know \( P(A\cap B)=\frac{68}{552}\) and \( P(B)=\frac{272}{552}\)
So, \( P(A|B)=\frac{\frac{68}{552}}{\frac{272}{552}}=\frac{68}{272}=\frac{68\div68}{272\div68}=\frac{1}{4}=0.25\)

Since \( P(A|B) = P(A)=0.25\), events A and B are independent.

Answer:

A and B are independent events because \( P(A|B)=P(A)\)