Question

the venn - diagram below shows the 17 students in ms. howards 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 garen, shen, and nicole 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 number of students in Chess Club

Count the number of students in circle A (Chess Club). There are 7 students in the Chess - Club. The total number of students is 17. So, $P(A)=\frac{7}{17}$.

Step2: Count number of students in Science Club

Count the number of students in circle B (Science Club). There are 8 students in the Science - Club. So, $P(B)=\frac{8}{17}$.

Step3: Count number of students in both clubs

The number of students in both the Chess Club and the Science Club (the intersection) is 1 (Lucy). So, $P(A\cap B)=\frac{1}{17}$.

Step4: Use conditional - probability formula

The formula for conditional probability is $P(B|A)=\frac{P(A\cap B)}{P(A)}$. Since $P(A\cap B)=\frac{1}{17}$ and $P(A)=\frac{7}{17}$, then $P(B|A)=\frac{\frac{1}{17}}{\frac{7}{17}}=\frac{1}{7}$.

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

We know that $P(A)=\frac{7}{17}$ and $P(B|A)=\frac{1}{7}$. So, $P(A)\cdot P(B|A)=\frac{7}{17}\times\frac{1}{7}=\frac{1}{17}$.

Answer:

$P(A)=\frac{7}{17}$
$P(B)=\frac{8}{17}$
$P(A\cap B)=\frac{1}{17}$
$P(B|A)=\frac{1}{7}$
$P(A)\cdot P(B|A)=\frac{1}{17}$