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

Explanation:

Step1: Identify probabilities

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

Step2: Apply the addition - rule of probability

The formula for \(P(A\cup B)\) (probability of \(A\) or \(B\) or both) is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).
So \(P = \frac{11}{40}+\frac{1}{4}-\frac{1}{20}\).

Answer:

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