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

Explanation:

Step1: Calculate probability of taking Consumer Education

The probability of taking Consumer Education, $P(C)$, is $\frac{22}{80}=\frac{11}{40}$ since there are 22 students taking it out of 80 total students.

Step2: Calculate probability of taking French

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

Step3: Calculate probability of taking both

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

Step4: Use the addition - rule of probability

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

Answer:

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