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}{20}) (p=\frac{11}{40}+\frac{1}{4}) (p=\frac{11}{40}+\frac{1}{4}-\frac{1}{10})

Explanation:

Step1: Calculate probability of taking Consumer Education

The probability of taking Consumer Education, $P(C)=\frac{22}{80}=\frac{11}{40}$.

Step2: Calculate probability of taking French

The probability of taking French, $P(F)=\frac{20}{80}=\frac{1}{4}$.

Step3: Calculate probability of taking both

The probability of taking both, $P(C\cap F)=\frac{4}{80}=\frac{1}{20}$.

Step4: Use the addition - rule for probability

The formula for $P(C\cup F)$ (probability of taking Consumer Education, French, or both) is $P(C\cup F)=P(C)+P(F)-P(C\cap F)$. Substituting the values we get $P=\frac{11}{40}+\frac{1}{4}-\frac{1}{20}$.

Answer:

$P = \frac{11}{40}+\frac{1}{4}-\frac{1}{20}$