Question

the venn - diagram below shows the 11 students in ms. daviss 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 hans and diane 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(b|a)= p(a) - p(b|a)=

Explanation:

Step1: Count total students

There are 11 students in total.

Step2: Find \(P(A)\)

There are 4 students in the Tennis - Club (A). So \(P(A)=\frac{4}{11}\).

Step3: Find \(P(B)\)

There are 6 students in the Soccer - Club (B). So \(P(B)=\frac{6}{11}\).

Step4: Find \(P(A\cap B)\)

There is 1 student in both clubs. So \(P(A\cap B)=\frac{1}{11}\).

Step5: Find \(P(B|A)\)

Use the formula \(P(B|A)=\frac{P(A\cap B)}{P(A)}\). Substitute \(P(A\cap B)=\frac{1}{11}\) and \(P(A)=\frac{4}{11}\), then \(P(B|A)=\frac{\frac{1}{11}}{\frac{4}{11}}=\frac{1}{4}\).

Step6: Find \(P(A)\cdot P(B|A)\)

\(P(A)\cdot P(B|A)=\frac{4}{11}\times\frac{1}{4}=\frac{1}{11}\).

Answer:

\(P(A)=\frac{4}{11}\)
\(P(B)=\frac{6}{11}\)
\(P(A\cap B)=\frac{1}{11}\)
\(P(B|A)=\frac{1}{4}\)
\(P(A)\cdot P(B|A)=\frac{1}{11}\)