QuestionA university surveys 50 undergraduates. Given $P(A \mid B)=\frac{15}{35}$ and $P(B \mid A)=\frac{15}{25}$ , fill in the Venn diagram for events $A$ , $B$ , $A \cap B$ , and neither.

Studdy Solution

STEP 1

Assumptions1. The total number of students surveyed is50. . Event A represents students that like to take online classes.
3. Event B represents first year students.
4. The conditional probability $(A \mid B)$ represents the probability of a student liking online classes given that they are a first year student.
5. The conditional probability $(B \mid A)$ represents the probability of a student being a first year student given that they like online classes.
6. The number of students who like online classes is25.
7. The number of first year students is35.

STEP 2

First, we need to find the number of students that belong to both events A and B. This can be found using the conditional probability $(A \mid B)$ which is the probability of a student liking online classes given that they are a first year student. $Number\, of\, students\, in\, A\, and\, B =(A \mid B) \times Number\, of\, students\, in\, B$

STEP 3

Now, plug in the given values for the conditional probability $(A \mid B)$ and the number of students in B to calculate the number of students in both A and B.
$Number\, of\, students\, in\, A\, and\, B = \frac{15}{35} \times35$

STEP 4

Calculate the number of students in both A and B.
$Number\, of\, students\, in\, A\, and\, B = \frac{15}{35} \times35 =15$

STEP 5

Now, we can find the number of students that belong only to event A (like online classes but are not first year students). This can be found by subtracting the number of students in both A and B from the total number of students in A.
$Number\, of\, students\, in\, A\, only = Number\, of\, students\, in\, A - Number\, of\, students\, in\, A\, and\, B$

STEP 6

Plug in the values for the number of students in A and the number of students in both A and B to calculate the number of students in A only.
$Number\, of\, students\, in\, A\, only =25 -15$

STEP 7

Calculate the number of students in A only.
$Number\, of\, students\, in\, A\, only =25 -15 =10$

STEP 8

Similarly, we can find the number of students that belong only to event B (first year students who do not like online classes). This can be found by subtracting the number of students in both A and B from the total number of students in B.
$Number\, of\, students\, in\, B\, only = Number\, of\, students\, in\, B - Number\, of\, students\, in\, A\, and\, B$

STEP 9

Plug in the values for the number of students in B and the number of students in both A and B to calculate the number of students in B only.
$Number\, of\, students\, in\, B\, only =35 -15$

STEP 10

Calculate the number of students in B only.
$Number\, of\, students\, in\, B\, only =35 -15 =20$

STEP 11

Finally, we can find the number of students that do not belong to either event A or B. This can be found by subtracting the number of students in A, B and both A and B from the total number of students.
$Number\, of\, students\, not\, in\, A\, or\, B = Total\, number\, of\, students - Number\, of\, students\, in\, A - Number\, of\, students\, in\, B + Number\, of\, students\, in\, A\, and\, B$

STEP 12

Plug in the values for the total number of students, the number of students in A, the number of students in B and the number of students in both A and B to calculate the number of students not in A or B.
$Number\, of\, students\, not\, in\, A\, or\, B =50 -25 -35 +15$

STEP 13

Calculate the number of students not in A or B.
$Number\, of\, students\, not\, in\, A\, or\, B =50 -25 -35 +15 =5$ So, the Venn diagram would be filled as follows-10 students like online classes only (Event A) -20 students are first year students only (Event B) -15 students are both first year students and like online classes (Event A and B) -5 students neither like online classes nor are they first year students (Not A or B)

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