Question

unit 1 test form a: probability

1. the results of a fruit preference survey are listed in the venn diagram below. peoples’ preferences were limited to either apples, bananas, or cantaloupe.

a. how many total people were surveyed?
find the probabilities of each of the following:
b. p(a∪b)
c. p(b∩c)
d. p(c)

1. 150 college students were interviewed. 85 of them were registered for a math class, 70 of them were registered for an english class and 45 of them were registered for neither.

a. create a venn diagram to represent the situation:
b. how many students were registered for both math and english class?

Explanation:

Step1: Calculate total number of people surveyed in question 14

Add all the numbers in the Venn - Diagram: $12 + 2+9 + 4+3 + 5+11=46$.

Step2: Calculate $P(A\cup B)$ in question 14

The elements in $A\cup B$ are the sum of elements in $A$ and $B$ and their intersection. $n(A\cup B)=12 + 2+9 + 4+3 + 5=35$. So $P(A\cup B)=\frac{35}{46}$.

Step3: Calculate $P(B\cap C)$ in question 14

The elements in $B\cap C$ are $3$ and $5$. So $n(B\cap C)=3 + 5 = 8$, and $P(B\cap C)=\frac{8}{46}=\frac{4}{23}$.

Step4: Calculate $P(C')$ in question 14

$C'$ means the complement of $C$. The elements not in $C$ are $12,2,9$. So $n(C')=12 + 2+9 + 3+5=35$, and $P(C')=\frac{35}{46}$.

Step5: Solve question 15b

Let $M$ be the set of students in Math class and $E$ be the set of students in English class. The total number of students is $N = 150$, and the number of students in neither is $45$. So the number of students in either Math or English or both is $150 - 45=105$. We know that $n(M)=85$ and $n(E)=70$. Using the formula $n(M\cup E)=n(M)+n(E)-n(M\cap E)$, we substitute $n(M\cup E) = 105$, $n(M)=85$ and $n(E)=70$. Then $105=85 + 70 - n(M\cap E)$. Solving for $n(M\cap E)$ gives $n(M\cap E)=85 + 70-105 = 50$.

Answer:

14a. 46
14b. $\frac{35}{46}$
14c. $\frac{4}{23}$
14d. $\frac{35}{46}$
15b. 50