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 = 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

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

Answer

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})