Question

in a freshman high school class of 80 students, 22 students take consumer education, 20 students take french, and 4 students take both. which equation can be used to find the probability, p, that a randomly selected student from this class takes consumer education, french, or both?
$p = \frac{11}{40} + \frac{1}{4} - \frac{1}{20}$
$p = \frac{11}{40} + \frac{1}{4} - \frac{1}{10}$
$p = \frac{11}{40} + \frac{1}{4} + \frac{1}{20}$
$p = \frac{11}{40} + \frac{1}{4}$

Explanation:

Step1: Find probabilities of each event

Total students = 80.
Probability of Consumer Education (CE): \( \frac{22}{80} = \frac{11}{40} \).
Probability of French (F): \( \frac{20}{80} = \frac{1}{4} \).
Probability of both (CE and F): \( \frac{4}{80} = \frac{1}{20} \).

Step2: Apply inclusion - exclusion principle

For two events \( A \) and \( B \), \( P(A \cup B)=P(A)+P(B)-P(A \cap B) \).
Here, \( A \) is CE, \( B \) is F. So \( P(\text{CE or French}) = P(\text{CE})+P(\text{French})-P(\text{CE and French}) \).
Substituting values: \( P=\frac{11}{40}+\frac{1}{4}-\frac{1}{20} \).

Answer:

\( P = \frac{11}{40}+\frac{1}{4}-\frac{1}{20} \) (the first option)