in a freshman high school class of 80 students, 22 students take consumer education, 20 students take…

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}

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}

Answer

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})