Question

the diagram below shows the memberships for the chess club and the science club for the 13 students in ms. morgans class. a student from the class is randomly selected. let a denote the event \the student is in the chess club.\ let b denote the event \the student is in the science club.\ the outcomes for the event a are listed in the circle on the left. the outcomes for the event b are listed in the circle on the right. note that diane, jane, and austin are outside the circles since they are not members of either club. (a) find the probabilities of the events below. write each answer as a single fraction. p(a)=□ p(b)=□ p(a and b)=□ p(a | b)=□ p(b)·p(a | b)=□

Explanation:

Step1: Calculate \(P(A)\)

Count the number of students in the Science - Club (event \(A\)). There are 7 students in the Science - Club. The total number of students is 13. So \(P(A)=\frac{7}{13}\).

Step2: Calculate \(P(B)\)

Count the number of students in the Chess - Club (event \(B\)). There are 8 students in the Chess - Club. So \(P(B)=\frac{8}{13}\).

Step3: Calculate \(P(A\cap B)\)

Count the number of students in both the Science - Club and the Chess - Club. There are 4 students in both clubs. So \(P(A\cap B)=\frac{4}{13}\).

Step4: Calculate \(P(A|B)\)

Use the formula \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Substitute \(P(A\cap B)=\frac{4}{13}\) and \(P(B)=\frac{8}{13}\) into the formula. \(P(A|B)=\frac{\frac{4}{13}}{\frac{8}{13}}=\frac{4}{8}=\frac{1}{2}\).

Step5: Calculate \(P(B)\cdot P(A|B)\)

We know \(P(B)=\frac{8}{13}\) and \(P(A|B)=\frac{1}{2}\). Then \(P(B)\cdot P(A|B)=\frac{8}{13}\times\frac{1}{2}=\frac{4}{13}\).

Answer:

\(P(A)=\frac{7}{13}\), \(P(B)=\frac{8}{13}\), \(P(A\cap B)=\frac{4}{13}\), \(P(A|B)=\frac{1}{2}\), \(P(B)\cdot P(A|B)=\frac{4}{13}\)