Question

the activity preference of teachers and librarians is given in the table.

	teachers	librarians
books	4	9
plays	2	15

based on the information in the table, which of the following is an example of independent events.
p(librarian and chooses book)
p(librarian and chooses movie)
p(teacher and chooses play)
p(teacher and chooses book)

Explanation:

Step1: Calculate total number of people

Total = (3 + 4+ 2)+(12 + 9+ 15)=45

Step2: Recall the concept of independent - events formula

For two events A and B to be independent, \(P(A\cap B)=P(A)\times P(B)\). In the context of this problem, we need to check if the probability of being a certain profession and having a certain preference can be written as the product of the probability of being that profession and the probability of having that preference. But we can also use the intuitive understanding that if the preference is not affected by the profession, we are looking for a situation where the proportion of the preference within a profession is the same as the proportion of that preference overall.
Let's calculate the probabilities for each option:

Option 1: \(P(\text{librarian and chooses book})\)

The number of librarians who choose books is 9. \(P(\text{librarian})=\frac{12 + 9+ 15}{45}=\frac{36}{45}=\frac{4}{5}\), \(P(\text{chooses book})=\frac{4 + 9}{45}=\frac{13}{45}\), \(P(\text{librarian and chooses book})=\frac{9}{45}\), and \(P(\text{librarian})\times P(\text{chooses book})=\frac{4}{5}\times\frac{13}{45}=\frac{52}{225}
eq\frac{9}{45}\)

Option 2: \(P(\text{librarian and chooses movie})\)

The number of librarians who choose movies is 12. \(P(\text{librarian})=\frac{36}{45}\), \(P(\text{chooses movie})=\frac{3 + 12}{45}=\frac{15}{45}\), \(P(\text{librarian and chooses movie})=\frac{12}{45}\), and \(P(\text{librarian})\times P(\text{chooses movie})=\frac{36}{45}\times\frac{15}{45}=\frac{36\times15}{45\times45}=\frac{540}{2025}=\frac{12}{45}\)

Option 3: \(P(\text{teacher and chooses play})\)

The number of teachers who choose plays is 2. \(P(\text{teacher})=\frac{3 + 4+ 2}{45}=\frac{9}{45}\), \(P(\text{chooses play})=\frac{2+15}{45}=\frac{17}{45}\), \(P(\text{teacher and chooses play})=\frac{2}{45}\), and \(P(\text{teacher})\times P(\text{chooses play})=\frac{9}{45}\times\frac{17}{45}=\frac{153}{2025}
eq\frac{2}{45}\)

Option 4: \(P(\text{teacher and chooses book})\)

The number of teachers who choose books is 4. \(P(\text{teacher})=\frac{9}{45}\), \(P(\text{chooses book})=\frac{13}{45}\), \(P(\text{teacher and chooses book})=\frac{4}{45}\), and \(P(\text{teacher})\times P(\text{chooses book})=\frac{9}{45}\times\frac{13}{45}=\frac{117}{2025}
eq\frac{4}{45}\)

Answer:

P(librarian and chooses movie)