Question

the venn - diagram below shows the 11 students in mr. washingtons class. the diagram shows the memberships for the chess club and the spanish club. note that \leila\, \david\, \yolanda\, and \ivan\ are outside the circles since they are not members of either club. one 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 spanish club.\ (a) find the probabilities of the events below. write each answer as a single fraction. (p(a)=square) (p(b)=square) (p(a or b)=square) (p(a and b)=square) (p(a)+p(b)-p(a and b)=square) (b) select the probability that is equal to (p(a)+p(b)-p(a and b)). (p(a or b)) (p(a)) (p(a and b))

Explanation:

Step1: Count members of Chess Club

There are 5 students (Diane, Rita, Christine, Hans, Pablo) in Chess Club. Total students are 11. So $P(A)=\frac{5}{11}$.

Step2: Count members of Spanish Club

There are 4 students (Pablo, Michael, Mary, Michael) in Spanish Club. So $P(B)=\frac{4}{11}$.

Step3: Count members of A or B

There are 7 students (Diane, Rita, Christine, Hans, Pablo, Michael, Mary) in either Chess or Spanish Club. So $P(A\ or\ B)=\frac{7}{11}$.

Step4: Count members of A and B

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

Step5: Calculate $P(A)+P(B)-P(A\ and\ B)$

$P(A)+P(B)-P(A\ and\ B)=\frac{5}{11}+\frac{4}{11}-\frac{1}{11}=\frac{5 + 4-1}{11}=\frac{8}{11}$.

Answer:

$P(A)=\frac{5}{11}$
$P(B)=\frac{4}{11}$
$P(A\ or\ B)=\frac{7}{11}$
$P(A\ and\ B)=\frac{1}{11}$
$P(A)+P(B)-P(A\ and\ B)=\frac{8}{11}$
(b) A. $P(A\ or\ B)$