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 = 11/40 + 1/4 - 1/10
p = 11/40 + 1/4
p = 11/40 + 1/4 + 1/20
p = 11/40 + 1/4 - 1/20

Explanation:

Step1: Recall probability formula

Use $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Let $A$ be taking Consumer - Education and $B$ be taking French. $P(A)=\frac{22}{80}=\frac{11}{40}$, $P(B)=\frac{20}{80}=\frac{1}{4}$, $P(A\cap B)=\frac{4}{80}=\frac{1}{20}$.

Step2: Substitute values

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

Answer:

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