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

Explanation:

Answer:

Let \(A\) be the event that a student takes Consumer - Education and \(B\) be the event that a student takes French. We know \(n = 80\), \(n(A)=20\), \(n(B) = 22\), \(n(A\cap B)=4\). The probability of an event \(E\) is \(P(E)=\frac{n(E)}{n(S)}\), so \(P(A)=\frac{20}{80}=\frac{1}{4}\), \(P(B)=\frac{22}{80}=\frac{11}{40}\), \(P(A\cap B)=\frac{4}{80}=\frac{1}{20}\).

The formula for \(P(A\cup B)\) is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Substituting the values we get \(P(A\cup B)=\frac{11}{40}+\frac{1}{4}-\frac{1}{20}\). So the correct equation is \(P = \frac{11}{40}+\frac{1}{4}-\frac{1}{20}\)