Question

the venn diagram below shows the 14 students in ms. harriss class. the diagram shows the memberships for the tennis club and the soccer club. a student from the class is randomly selected. let a denote the event \the student is in the tennis club.\ let b denote the event \the student is in the soccer 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 scott is outside the circles since he is not a member 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 \(P(A)\)

The number of students in event \(A\) (Tennis Club) is \(6 + 2=8\). The total number of students is \(14\). So \(P(A)=\frac{8}{14}=\frac{4}{7}\).

Step2: Calculate \(P(B)\)

The number of students in event \(B\) (Soccer Club) is \(7+ 2=9\). The total number of students is \(14\). So \(P(B)=\frac{9}{14}\).

Step3: Calculate \(P(A\text{ and }B)\)

The number of students in both \(A\) and \(B\) is \(2\). The total number of students is \(14\). So \(P(A\text{ and }B)=\frac{2}{14}=\frac{1}{7}\).

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

By the formula \(P(A|B)=\frac{P(A\text{ and }B)}{P(B)}\), substituting \(P(A\text{ and }B)=\frac{1}{7}\) and \(P(B)=\frac{9}{14}\), we get \(P(A|B)=\frac{\frac{1}{7}}{\frac{9}{14}}=\frac{1}{7}\times\frac{14}{9}=\frac{2}{9}\).

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

\(P(B)\cdot P(A|B)=\frac{9}{14}\times\frac{2}{9}=\frac{1}{7}\).

Answer:

\(P(A)=\frac{4}{7}\)
\(P(B)=\frac{9}{14}\)
\(P(A\text{ and }B)=\frac{1}{7}\)
\(P(A|B)=\frac{2}{9}\)
\(P(B)\cdot P(A|B)=\frac{1}{7}\)