Question

the venn - diagram below shows the 11 students in ms. murphys class. the diagram shows the memberships for the chess club and the science club. 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 justin 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(b | a)=□ p(a)·p(b | a)=□

Explanation:

Step1: Calculate \(P(A)\)

The number of students in the Science - Club (event \(A\)) is 6 (Lena, Debra, Rita, Austin, Linda, Bob). The total number of students is 11. So \(P(A)=\frac{6}{11}\).

Step2: Calculate \(P(B)\)

The number of students in the Chess - Club (event \(B\)) is 5 (Reuben, Lena, Debra, Rita, Isabel). So \(P(B)=\frac{5}{11}\).

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

The number of students in both clubs (\(A\cap B\)) is 3 (Lena, Debra, Rita). So \(P(A\cap B)=\frac{3}{11}\).

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

The formula for conditional probability is \(P(B|A)=\frac{P(A\cap B)}{P(A)}\). Substituting the values we found: \(P(B|A)=\frac{\frac{3}{11}}{\frac{6}{11}}=\frac{3}{6}=\frac{1}{2}\).

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

\(P(A)\cdot P(B|A)=\frac{6}{11}\times\frac{1}{2}=\frac{3}{11}\).

Answer:

\(P(A)=\frac{6}{11}\)
\(P(B)=\frac{5}{11}\)
\(P(A\ and\ B)=\frac{3}{11}\)
\(P(B|A)=\frac{1}{2}\)
\(P(A)\cdot P(B|A)=\frac{3}{11}\)