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{22}{80}+\frac{20}{80} p = \frac{22}{80}+\frac{20}{80}-\frac{4}{80} p = \frac{22}{80}+\frac{4}{80}-\frac{20}{80} p = \frac{22}{80}+\frac{4}{80}+\frac{20}{80}

Explanation:

Answer:

Let \(A\) be the set of students taking Consumer - Education and \(B\) be the set of students taking French. We know \(n(A)=22\), \(n(B) = 20\), \(n(A\cap B)=4\) and \(n(S)=80\) (where \(S\) is the total number of students in the class).

The formula for \(P(A\cup B)\) (probability of \(A\) or \(B\) or both) is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).

Since \(P(A)=\frac{n(A)}{n(S)}=\frac{22}{80}\), \(P(B)=\frac{n(B)}{n(S)}=\frac{20}{80}\) and \(P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{4}{80}\)

The equation is \(P=\frac{22}{80}+\frac{20}{80}-\frac{4}{80}\)

So the correct option is \(P=\frac{22}{80}+\frac{20}{80}-\frac{4}{80}\)